Hyperbola standard form

Hyperbola standard form DEFAULT

8.3: The Hyperbola

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Learning Objectives

  • Locate a hyperbola’s vertices and foci.
  • Write equations of hyperbolas in standard form.
  • Graph hyperbolas centered at the origin.
  • Graph hyperbolas not centered at the origin.
  • Solve applied problems involving hyperbolas.

What do paths of comets, supersonic booms, ancient Grecian pillars, and natural draft cooling towers have in common? They can all be modeled by the same type of conic. For instance, when something moves faster than the speed of sound, a shock wave in the form of a cone is created. A portion of a conic is formed when the wave intersects the ground, resulting in a sonic boom (Figure \(\PageIndex{1}\)).

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Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. The crack of a whip occurs because the tip is exceeding the speed of sound. The bullets shot from many firearms also break the sound barrier, although the bang of the gun usually supersedes the sound of the sonic boom.

Locating the Vertices and Foci of a Hyperbola

In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other (Figure \(\PageIndex{2}\)).

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Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points \((x,y)\) in a plane such that the difference of the distances between \((x,y)\) and the foci is a positive constant.

Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances.

As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle (Figure \(\PageIndex{3}\)).

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In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the \(x\)- and \(y\)-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin.

Deriving the Equation of an Ellipse Centered at the Origin

Let \((−c,0)\) and \((c,0)\) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points \((x,y)\) such that the difference of the distances from \((x,y)\) to the foci is constant. See Figure \(\PageIndex{4}\).

A horizontal hyperbola in the x y coordinate system centered at (0, 0) with Vertices at (negative a, 0) and (a, 0) and Foci at (negative c, 0) and (c, 0), with lines of length d1 and d2 connecting a point on the right branch of the hyperbola to the foci.

If \((a,0)\) is a vertex of the hyperbola, the distance from \((−c,0)\) to \((a,0)\) is \(a−(−c)=a+c\). The distance from \((c,0)\) to \((a,0)\) is \(c−a\). The sum of the distances from the foci to the vertex is

\((a+c)−(c−a)=2a\)

If \((x,y)\) is a point on the hyperbola, we can define the following variables:

\(d_2=\) the distance from \((−c,0)\) to \((x,y)\)

\(d_1=\) the distance from \((c,0)\) to \((x,y)\)

By definition of a hyperbola, \(d_2−d_1\) is constant for any point \((x,y)\) on the hyperbola. We know that the difference of these distances is \(2a\) for the vertex \((a,0)\). It follows that \(d_2−d_1=2a\) for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses.

\[\begin{align*} d_2-d_1&=2a\\ \sqrt{{(x-(-c))}^2+{(y-0)}^2}-\sqrt{{(x-c)}^2+{(y-0)}^2}&=2a\qquad \text{Distance Formula}\\ \sqrt{{(x+c)}^2+y^2}-\sqrt{{(x-c)}^2+y^2}&=2a\qquad \text{Simplify expressions.}\\ \sqrt{{(x+c)}^2+y^2}&=2a+\sqrt{{(x-c)}^2+y^2}\qquad \text{Move radical to opposite side.}\\ {(x+c)}^2+y^2&={(2a+\sqrt{{(x-c)}^2+y^2})}^2\qquad \text{Square both sides.}\\ x^2+2cx+c^2+y^2&=4a^2+4a\sqrt{{(x-c)}^2+y^2}+{(x-c)}^2+y^2\qquad \text{Expand the squares.}\\ x^2+2cx+c^2+y^2&=4a^2+4a\sqrt{{(x-c)}^2+y^2}+x^2-2cx+c^2+y^2\qquad \text{Expand remaining square.}\\ 2cx&=4a^2+4a\sqrt{{(x-c)}^2+y^2}-2cx\qquad \text{Combine like terms.}\\ 4cx-4a^2&=4a\sqrt{{(x-c)}^2+y^2}\qquad \text{Isolate the radical.}\\ cx-a^2&=a\sqrt{{(x-c)}^2+y^2}\qquad \text{Divide by 4.}\\ {(cx-a^2)}^2&=a^2{\left[\sqrt{{(x-c)}^2+y^2}\right]}^2\qquad \text{Square both sides.}\\ c^2x^2-2a^2cx+a^4&=a^2(x^2-2cx+c^2+y^2)\qquad \text{Expand the squares.}\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\qquad \text{Distribute } a^2\\ a^4+c^2x^2&=a^2x^2+a^2c^2+a^2y^2\qquad \text{Combine like terms.}\\ c^2x^2-a^2x^2-a^2y^2&=a^2c^2-a^4\qquad \text{Rearrange terms.}\\ x^2(c^2-a^2)-a^2y^2&=a^2(c^2-a^2)\qquad \text{Factor common terms.}\\ x^2b^2-a^2y^2&=a^2b^2\qquad \text{Set } b^2=c^2−a^2\\. \dfrac{x^2b^2}{a^2b^2}-\dfrac{a^2y^2}{a^2b^2}&=\dfrac{a^2b^2}{a^2b^2}\qquad \text{Divide both sides by } a^2b^2\\ \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}&=1\\ \end{align*}\]

This equation defines a hyperbola centered at the origin with vertices \((\pm a,0)\) and co-vertices \((0,\pm b)\).

STANDARD FORMS OF THE EQUATION OF A HYPERBOLA WITH CENTER \((0,0)\)

The standard form of the equation of a hyperbola with center \((0,0)\) and transverse axis on the \(x\)-axis is

\(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\)

where

  • the length of the transverse axis is \(2a\)
  • the coordinates of the vertices are \((\pm a,0)\)
  • the length of the conjugate axis is \(2b\)
  • the coordinates of the co-vertices are \((0,\pm b)\)
  • the distance between the foci is \(2c\), where \(c^2=a^2+b^2\)
  • the coordinates of the foci are \((\pm c,0)\)
  • the equations of the asymptotes are \(y=\pm \dfrac{b}{a}x\)

See Figure \(\PageIndex{5a}\).

The standard form of the equation of a hyperbola with center \((0,0)\) and transverse axis on the \(y\)-axis is

\(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\)

where

  • the length of the transverse axis is \(2a\)
  • the coordinates of the vertices are \((0,\pm a)\)
  • the length of the conjugate axis is \(2b\)
  • the coordinates of the co-vertices are \((\pm b,0)\)
  • the distance between the foci is \(2c\), where \(c^2=a^2+b^2\)
  • the coordinates of the foci are \((0,\pm c)\)
  • the equations of the asymptotes are \(y=\pm \dfrac{a}{b}x\)

See Figure \(\PageIndex{5b}\).

Note that the vertices, co-vertices, and foci are related by the equation \(c^2=a^2+b^2\). When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

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How to: Given the equation of a hyperbola in standard form, locate its vertices and foci

  1. Determine whether the transverse axis lies on the \(x\)- or \(y\)-axis. Notice that \(a^2\) is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.
    • If the equation has the form \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\), then the transverse axis lies on the \(x\)-axis. The vertices are located at \((\pm a,0)\), and the foci are located at \((\pm c,0)\).
    • If the equation has the form \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\), then the transverse axis lies on the \(y\)-axis. The vertices are located at \((0,\pm a)\), and the foci are located at \((0,\pm c)\).
  2. Solve for \(a\) using the equation \(a=\sqrt{a^2}\).
  3. Solve for \(c\) using the equation \(c=\sqrt{a^2+b^2}\).

Example \(\PageIndex{1}\): Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation \(\dfrac{y^2}{49}−\dfrac{x^2}{32}=1\).

Solution

The equation has the form \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\), so the transverse axis lies on the \(y\)-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set \(x=0\), and solve for \(y\).

\[\begin{align*} 1&=\dfrac{y^2}{49}-\dfrac{x^2}{32}\\ 1&=\dfrac{y^2}{49}-\dfrac{0^2}{32}\\ 1&=\dfrac{y^2}{49}\\ y^2&=49\\ y&=\pm \sqrt{49}\\ &=\pm 7 \end{align*}\]

The foci are located at \((0,\pm c)\). Solving for \(c\),

\[\begin{align*} c&=\sqrt{a^2+b^2}\\ &=\sqrt{49+32}\\ &=\sqrt{81}\\ &=9 \end{align*}\]

Therefore, the vertices are located at \((0,\pm 7)\), and the foci are located at \((0,9)\).

Exercise \(\PageIndex{1}\)

Identify the vertices and foci of the hyperbola with equation \(\dfrac{x^2}{9}−\dfrac{y^2}{25}=1\).

Answer

Vertices: \((\pm 3,0)\); Foci: \((\pm \sqrt{34},0)\)

Writing Equations of Hyperbolas in Standard Form

Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.

Hyperbolas Centered at the Origin

Reviewing the standard forms given for hyperbolas centered at \((0,0)\),we see that the vertices, co-vertices, and foci are related by the equation \(c^2=a^2+b^2\). Note that this equation can also be rewritten as \(b^2=c^2−a^2\). This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.

How to: Given the vertices and foci of a hyperbola centered at \((0,0)\), write its equation in standard form

  1. Determine whether the transverse axis lies on the \(x\)- or \(y\)-axis.
    • If the given coordinates of the vertices and foci have the form \((\pm a,0)\) and \((\pm c,0)\), respectively, then the transverse axis is the \(x\)-axis. Use the standard form \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\).
    • If the given coordinates of the vertices and foci have the form \((0,\pm a)\) and \((0,\pm c)\), respectively, then the transverse axis is the \(y\)-axis. Use the standard form \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\).
  2. Find \(b^2\) using the equation \(b^2=c^2−a^2\).
  3. Substitute the values for \(a^2\) and \(b^2\) into the standard form of the equation determined in Step 1.

Example \(\PageIndex{2}\): Finding the Equation of a Hyperbola Centered at \((0,0)\) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices \((\pm 6,0)\) and foci \((\pm 2\sqrt{10},0)\)?

Solution

The vertices and foci are on the \(x\)-axis. Thus, the equation for the hyperbola will have the form \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\).

The vertices are \((\pm 6,0)\), so \(a=6\) and \(a^2=36\).

The foci are \((\pm 2\sqrt{10},0)\), so \(c=2\sqrt{10}\) and \(c^2=40\).

Solving for \(b^2\), we have

\[\begin{align*} b^2&=c^2-a^2\\ b^2&=40-36\qquad \text{Substitute for } c^2 \text{ and } a^2\\ b^2&=4\qquad \text{Subtract.} \end{align*}\]

Finally, we substitute \(a^2=36\) and \(b^2=4\) into the standard form of the equation, \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\). The equation of the hyperbola is \(\dfrac{x^2}{36}−\dfrac{y^2}{4}=1\), as shown in Figure \(\PageIndex{6}\).

A horizontal hyperbola centered at (0, 0) in the x y coordinate system with Vertices at (negative 6, 0) and (6, 0).

Exercise \(\PageIndex{2}\)

What is the standard form equation of the hyperbola that has vertices \((0,\pm 2)\) and foci \((0,\pm 2\sqrt{5})\)?

Answer

\(\dfrac{y^2}{4}−\dfrac{x^2}{16}=1\)

Hyperbolas Not Centered at the Origin

Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated \(h\) units horizontally and \(k\) units vertically, the center of the hyperbola will be \((h,k)\). This translation results in the standard form of the equation we saw previously, with \(x\) replaced by \((x−h)\) and \(y\) replaced by \((y−k)\).

STANDARD FORMS OF THE EQUATION OF A HYPERBOLA WITH CENTER \((H, K)\)

The standard form of the equation of a hyperbola with center \((h,k)\) and transverse axis parallel to the \(x\)-axis is

\[\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\]

where

  • the length of the transverse axis is \(2a\)
  • the coordinates of the vertices are \((h\pm a,k)\)
  • the length of the conjugate axis is \(2b\)
  • the coordinates of the co-vertices are \((h,k\pm b)\)
  • the distance between the foci is \(2c\), where \(c^2=a^2+b^2\)
  • the coordinates of the foci are \((h\pm c,k)\)

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is \(2a\) and its width is \(2b\). The slopes of the diagonals are \(\pm \dfrac{b}{a}\),and each diagonal passes through the center \((h,k)\). Using the point-slope formula, it is simple to show that the equations of the asymptotes are \(y=\pm \dfrac{b}{a}(x−h)+k\). See Figure \(\PageIndex{7a}\).

The standard form of the equation of a hyperbola with center \((h,k)\) and transverse axis parallel to the \(y\)-axis is

\[\dfrac{{(y−k)}^2}{a^2}−\dfrac{{(x−h)}^2}{b^2}=1\]

where

  • the length of the transverse axis is \(2a\)
  • the coordinates of the vertices are \((h,k\pm a)\)
  • the length of the conjugate axis is \(2b\)
  • the coordinates of the co-vertices are \((h\pm b,k)\)
  • the distance between the foci is \(2c\), where \(c^2=a^2+b^2\)
  • the coordinates of the foci are \((h,k\pm c)\)

Using the reasoning above, the equations of the asymptotes are \(y=\pm \dfrac{a}{b}(x−h)+k\). See Figure \(\PageIndex{7b}\).

This is a horizontal parabola opening to the right with Vertex (0, 0), Focus (6, 0), and Directrix x = negative 6. The Latus Rectum is shown, a vertical line passing through the Focus and terminating on the parabola at (6, 12) and (6, negative 12).

Like hyperbolas centered at the origin, hyperbolas centered at a point \((h,k)\) have vertices, co-vertices, and foci that are related by the equation \(c^2=a^2+b^2\). We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.

How to: Given the vertices and foci of a hyperbola centered at \((h,k)\),write its equation in standard form

  1. Determine whether the transverse axis is parallel to the \(x\)- or \(y\)-axis.
    • If the \(y\)-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the \(x\)-axis. Use the standard form \(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\).
    • If the \(x\)-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the \(y\)-axis. Use the standard form \(\dfrac{{(y−k)}^2}{a^2}−\dfrac{{(x−h)}^2}{b^2}=1\).
  2. Identify the center of the hyperbola, \((h,k)\),using the midpoint formula and the given coordinates for the vertices.
  3. Find \(a^2\) by solving for the length of the transverse axis, \(2a\), which is the distance between the given vertices.
  4. Find \(c^2\) using \(h\) and \(k\) found in Step 2 along with the given coordinates for the foci.
  5. Solve for \(b^2\) using the equation \(b^2=c^2−a^2\).
  6. Substitute the values for \(h\), \(k\), \(a^2\), and \(b^2\) into the standard form of the equation determined in Step 1.

Example \(\PageIndex{3}\): Finding the Equation of a Hyperbola Centered at \((h, k)\) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices at \((0,−2)\) and \((6,−2)\) and foci at \((−2,−2)\) and \((8,−2)\)?

Solution

The \(y\)-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the \(x\)-axis. Thus, the equation of the hyperbola will have the form

\(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\)

First, we identify the center, \((h,k)\). The center is halfway between the vertices \((0,−2)\) and \((6,−2)\). Applying the midpoint formula, we have

\((h,k)=(\dfrac{0+6}{2},\dfrac{−2+(−2)}{2})=(3,−2)\)

Next, we find \(a^2\). The length of the transverse axis, \(2a\),is bounded by the vertices. So, we can find \(a^2\) by finding the distance between the \(x\)-coordinates of the vertices.

\[\begin{align*} 2a&=| 0-6 |\\ 2a&=6\\ a&=3\\ a^2&=9 \end{align*}\]

Now we need to find \(c^2\). The coordinates of the foci are \((h\pm c,k)\). So \((h−c,k)=(−2,−2)\) and \((h+c,k)=(8,−2)\). We can use the \(x\)-coordinate from either of these points to solve for \(c\). Using the point \((8,−2)\), and substituting \(h=3\),

\[\begin{align*} h+c&=8\\ 3+c&=8\\ c&=5\\ c^2&=25 \end{align*}\]

Next, solve for \(b^2\) using the equation \(b^2=c^2−a^2\):

\[\begin{align*} b^2&=c^2-a^2\\ &=25-9\\ &=16 \end{align*}\]

Finally, substitute the values found for \(h\), \(k\), \(a^2\),and \(b^2\) into the standard form of the equation.

\(\dfrac{{(x−3)}^2}{9}−\dfrac{{(y+2)}^2}{16}=1\)

Exercise \(\PageIndex{3}\)

What is the standard form equation of the hyperbola that has vertices \((1,−2)\) and \((1,8)\) and foci \((1,−10)\) and \((1,16)\)?

Answer

\(\dfrac{{(y−3)}^2}{25}+\dfrac{{(x−1)}^2}{144}=1\)

Graphing Hyperbolas Centered at the Origin

When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\) for horizontal hyperbolas and the standard form \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\) for vertical hyperbolas.

How to: Given a standard form equation for a hyperbola centered at \((0,0)\), sketch the graph

  1. Determine which of the standard forms applies to the given equation.
  2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes.
    • If the equation is in the form \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\), then
      • the transverse axis is on the \(x\)-axis
      • the coordinates of the vertices are \((\pm a,0)\0
      • the coordinates of the co-vertices are \((0,\pm b)\)
      • the coordinates of the foci are \((\pm c,0)\)
      • the equations of the asymptotes are \(y=\pm \dfrac{b}{a}x\)
    • If the equation is in the form \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\), then
      • the transverse axis is on the \(y\)-axis
      • the coordinates of the vertices are \((0,\pm a)\)
      • the coordinates of the co-vertices are \((\pm b,0)\)
      • the coordinates of the foci are \((0,\pm c)\)
      • the equations of the asymptotes are \(y=\pm \dfrac{a}{b}x\)
  3. Solve for the coordinates of the foci using the equation \(c=\pm \sqrt{a^2+b^2}\).
  4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.

Example \(\PageIndex{4}\): Graphing a Hyperbola Centered at \((0,0)\) Given an Equation in Standard Form

Graph the hyperbola given by the equation \(\dfrac{y^2}{64}−\dfrac{x^2}{36}=1\). Identify and label the vertices, co-vertices, foci, and asymptotes.

Solution

The standard form that applies to the given equation is \(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\). Thus, the transverse axis is on the \(y\)-axis

The coordinates of the vertices are \((0,\pm a)=(0,\pm \sqrt{64})=(0,\pm 8)\)

The coordinates of the co-vertices are \((\pm b,0)=(\pm \sqrt{36}, 0)=(\pm 6,0)\)

The coordinates of the foci are \((0,\pm c)\), where \(c=\pm \sqrt{a^2+b^2}\). Solving for \(c\), we have

\(c=\pm \sqrt{a^2+b^2}=\pm \sqrt{64+36}=\pm \sqrt{100}=\pm 10\)

Therefore, the coordinates of the foci are \((0,\pm 10)\)

The equations of the asymptotes are \(y=\pm \dfrac{a}{b}x=\pm \dfrac{8}{6}x=\pm \dfrac{4}{3}x\)

Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in Figure \(\PageIndex{8}\).

A vertical hyperbola centered at (0, 0) in the x y coordinate system with Vertices at (0, 8) and (0, negative 8) and Foci at (0, negative 10) and (0, 10). Also shown are the slant asymptotes, y = (4/3)x and y = (negative 4/3)x. The points (negative 6, 0) (6, 0) and (0, 0) are labeled.

Exercise \(\PageIndex{4}\)

Graph the hyperbola given by the equation \(\dfrac{x^2}{144}−\dfrac{y^2}{81}=1\). Identify and label the vertices, co-vertices, foci, and asymptotes.

Answer

vertices: \((\pm 12,0)\); co-vertices: \((0,\pm 9)\); foci: \((\pm 15,0)\); asymptotes: \(y=\pm \dfrac{3}{4}x\);

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Graphing Hyperbolas Not Centered at the Origin

Graphing hyperbolas centered at a point \((h,k)\) other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms \(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\) for horizontal hyperbolas, and \(\dfrac{{(y−k)}^2}{a^2}−\dfrac{{(x−h)}^2}{b^2}=1\) for vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes.

How to: Given a general form for a hyperbola centered at \((h, k)\), sketch the graph

  1. Convert the general form to that standard form. Determine which of the standard forms applies to the given equation.
  2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes.
    1. If the equation is in the form \(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\), then
      • the transverse axis is parallel to the \(x\)-axis
      • the center is \((h,k)\)
      • the coordinates of the vertices are \((h\pm a,k)\)
      • the coordinates of the co-vertices are \((h,k\pm b)\)
      • the coordinates of the foci are \((h\pm c,k)\)
      • the equations of the asymptotes are \(y=\pm \dfrac{b}{a}(x−h)+k\)
    2. If the equation is in the form \(\dfrac{{(y−k)}^2}{a^2}−\dfrac{{(x−h)}^2}{b^2}=1\), then
      • the transverse axis is parallel to the \(y\)-axis
      • the center is \((h,k)\)
      • the coordinates of the vertices are \((h,k\pm a)\)
      • the coordinates of the co-vertices are \((h\pm b,k)\)
      • the coordinates of the foci are \((h,k\pm c)\)
      • the equations of the asymptotes are \(y=\pm \dfrac{a}{b}(x−h)+k\)
  3. Solve for the coordinates of the foci using the equation \(c=\pm \sqrt{a^2+b^2}\).
  4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.

Example \(\PageIndex{5}\): Graphing a Hyperbola Centered at \((h, k)\) Given an Equation in General Form

Graph the hyperbola given by the equation \(9x^2−4y^2−36x−40y−388=0\). Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Solution

Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation.

\((9x^2−36x)−(4y^2+40y)=388\)

Factor the leading coefficient of each expression.

\(9(x^2−4x)−4(y^2+10y)=388\)

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

\(9(x^2−4x+4)−4(y^2+10y+25)=388+36−100\)

Rewrite as perfect squares.

\(9{(x−2)}^2−4{(y+5)}^2=324\)

Divide both sides by the constant term to place the equation in standard form.

\(\dfrac{{(x−2)}^2}{36}−\dfrac{{(y+5)}^2}{81}=1\)

The standard form that applies to the given equation is \(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\), where \(a^2=36\) and \(b^2=81\),or \(a=6\) and \(b=9\). Thus, the transverse axis is parallel to the \(x\)-axis. It follows that:

the center of the ellipse is \((h,k)=(2,−5)\)

the coordinates of the vertices are \((h\pm a,k)=(2\pm 6,−5)\), or \((−4,−5)\) and \((8,−5)\)

the coordinates of the co-vertices are \((h,k\pm b)=(2,−5\pm 9)\), or \((2,−14)\) and \((2,4)\)

the coordinates of the foci are \((h\pm c,k)\), where \(c=\pm \sqrt{a^2+b^2}\). Solving for \(c\),we have

\(c=\pm \sqrt{36+81}=\pm \sqrt{117}=\pm 3\sqrt{13}\)

Therefore, the coordinates of the foci are \((2−3\sqrt{13},−5)\) and \((2+3\sqrt{13},−5)\).

The equations of the asymptotes are \(y=\pm \dfrac{b}{a}(x−h)+k=\pm \dfrac{3}{2}(x−2)−5\).

Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola, as shown in Figure \(\PageIndex{10}\).

A horizontal hyperbola centered at (2, negative 5) with Vertices at (negative 4, negative 5) and (8, 5) and Foci at (2 minus 3 square root of 13, negative 5) and (2 + 3 square root of 13, negative 5). Also shown are the slant asymptotes, y = (3/2) times (x minus 2) minus 5 and y = (negative 3/2)times (x minus 2) minus 5. The points (2, negative 14), (2, 4) and (0, 0) are labeled.

Exercise \(\PageIndex{5}\)

Graph the hyperbola given by the standard form of an equation \(\dfrac{{(y+4)}^2}{100}−\dfrac{{(x−3)}^2}{64}=1\). Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Answer

center: \((3,−4)\); vertices: \((3,−14)\) and \((3,6)\); co-vertices: \((−5,−4)\); and \((11,−4)\); foci: \((3,−4−2\sqrt{41})\) and \((3,−4+2\sqrt{41})\); asymptotes: \(y=\pm \dfrac{5}{4}(x−3)−4\)

alt

Solving Applied Problems Involving Hyperbolas

As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength (Figure \(\PageIndex{12}\)). For example, a \(500\)-foot tower can be made of a reinforced concrete shell only \(6\) or \(8\) inches wide!

alt

The first hyperbolic towers were designed in 1914 and were \(35\) meters high. Today, the tallest cooling towers are in France, standing a remarkable \(170\) meters tall. In Example \(\PageIndex{6}\) we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides.

Example \(\PageIndex{6}\): Solving Applied Problems Involving Hyperbolas

The design layout of a cooling tower is shown in Figure \(\PageIndex{13}\). The tower stands \(179.6\) meters tall. The diameter of the top is \(72\) meters. At their closest, the sides of the tower are \(60\) meters apart.

alt

Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

Solution

We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin: \(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\), where the branches of the hyperbola form the sides of the cooling tower. We must find the values of \(a^2\) and \(b^2\) to complete the model.

First, we find \(a^2\). Recall that the length of the transverse axis of a hyperbola is \(2a\). This length is represented by the distance where the sides are closest, which is given as \(65.3\) meters. So, \(2a=60\). Therefore, \(a=30\) and \(a^2=900\).

To solve for \(b^2\),we need to substitute for \(x\) and \(y\) in our equation using a known point. To do this, we can use the dimensions of the tower to find some point \((x,y)\) that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the \(y\)-axis bisects the tower, our \(x\)-value can be represented by the radius of the top, or \(36\) meters. The y-value is represented by the distance from the origin to the top, which is given as \(79.6\) meters. Therefore,

\[\begin{align*} \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}&=1\qquad \text{Standard form of horizontal hyperbola.}\\ b^2&=\dfrac{y^2}{\dfrac{x^2}{a^2}-1}\qquad \text{Isolate } b^2\\ &=\dfrac{{(79.6)}^2}{\dfrac{{(36)}^2}{900}-1}\qquad \text{Substitute for } a^2,\: x, \text{ and } y\\ &\approx 14400.3636\qquad \text{Round to four decimal places} \end{align*}\]

The sides of the tower can be modeled by the hyperbolic equation

\(\dfrac{x^2}{900}−\dfrac{y^2}{14400.3636}=1\),or \(\dfrac{x^2}{{30}^2}−\dfrac{y^2}{{120.0015}^2}=1\)

Exercise \(\PageIndex{6}\)

A design for a cooling tower project is shown in Figure \(\PageIndex{14}\). Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

Project design for a natural draft cooling tower. The overall height is 167.082 meters. The diameter at the top is 60 meters, and at their closest, 79.6 meters from the top, the sides are 60 meters apart.
Answer

The sides of the tower can be modeled by the hyperbolic equation. \(\dfrac{x^2}{400}−\dfrac{y^2}{3600}=1\) or \(\dfrac{x^2}{{20}^2}−\dfrac{y^2}{{60}^2}=1\).

Media

Access these online resources for additional instruction and practice with hyperbolas.

  • Conic Sections: The Hyperbola Part 1 of 2
  • Conic Sections: The Hyperbola Part 2 of 2
  • Graph a Hyperbola with Center at Origin
  • Graph a Hyperbola with Center not at Origin

Key Equations

Hyperbola, center at origin, transverse axis on x-axis\(\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1\)
Hyperbola, center at origin, transverse axis on y-axis\(\dfrac{y^2}{a^2}−\dfrac{x^2}{b^2}=1\)
Hyperbola, center at \((h,k)\),transverse axis parallel to x-axis\(\dfrac{{(x−h)}^2}{a^2}−\dfrac{{(y−k)}^2}{b^2}=1\)
Hyperbola, center at \((h,k)\),transverse axis parallel to y-axis\(\dfrac{{(y−k)}^2}{a^2}−\dfrac{{(x−h)}^2}{b^2}=1\)

Key Concepts

  • A hyperbola is the set of all points \((x,y)\) in a plane such that the difference of the distances between \((x,y)\) and the foci is a positive constant.
  • The standard form of a hyperbola can be used to locate its vertices and foci. See Example \(\PageIndex{1}\).
  • When given the coordinates of the foci and vertices of a hyperbola, we can write the equation of the hyperbola in standard form. See Example \(\PageIndex{2}\) and Example \(\PageIndex{3}\).
  • When given an equation for a hyperbola, we can identify its vertices, co-vertices, foci, asymptotes, and lengths and positions of the transverse and conjugate axes in order to graph the hyperbola. See Example \(\PageIndex{4}\) and Example \(\PageIndex{5}\).
  • Real-world situations can be modeled using the standard equations of hyperbolas. For instance, given the dimensions of a natural draft cooling tower, we can find a hyperbolic equation that models its sides. See Example \(\PageIndex{6}\).

Contributors and Attributions

Sours: https://math.libretexts.org/Bookshelves/Algebra/Map%3A_College_Algebra_(OpenStax)/08%3A_Analytic_Geometry/8.03%3A_The_Hyperbola

Learning Outcomes

  • Derive an equation for a hyperbola centered at the origin
  • Write an equation for a hyperbola centered at the origin
  • Solve an applied problem involving hyperbolas

In analytic geometry a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other.

A hyperbola

Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points [latex]\left(x,y\right)[/latex] in a plane such that the difference of the distances between [latex]\left(x,y\right)[/latex] and the foci is a positive constant.

Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances.

As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle.

Key features of the hyperbola

In this section we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x– and y-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin.

Standard Form of the Equation of a Hyperbola Centered at the Origin

Let [latex]\left(-c,0\right)[/latex] and [latex]\left(c,0\right)[/latex] be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points [latex]\left(x,y\right)[/latex] such that the difference of the distances from [latex]\left(x,y\right)[/latex] to the foci is constant.

If [latex]\left(a,0\right)[/latex] is a vertex of the hyperbola, the distance from [latex]\left(-c,0\right)[/latex] to [latex]\left(a,0\right)[/latex] is [latex]a-\left(-c\right)=a+c[/latex]. The distance from [latex]\left(c,0\right)[/latex] to [latex]\left(a,0\right)[/latex] is [latex]c-a[/latex]. The difference of the distances from the foci to the vertex is

[latex]\left(a+c\right)-\left(c-a\right)=2a[/latex]

If [latex]\left(x,y\right)[/latex] is a point on the hyperbola, we can define the following variables:

[latex]\begin{align}&{d}_{2}=\text{the distance from }\left(-c,0\right)\text{ to }\left(x,y\right)\\ &{d}_{1}=\text{the distance from }\left(c,0\right)\text{ to }\left(x,y\right)\end{align}[/latex]

By definition of a hyperbola, [latex]\lvert{d}_{2}-{d}_{1}\rvert[/latex] is constant for any point [latex]\left(x,y\right)[/latex] on the hyperbola. We know that the difference of these distances is [latex]2a[/latex] for the vertex [latex]\left(a,0\right)[/latex]. It follows that [latex]\lvert{d}_{2}-{d}_{1}\rvert=2a[/latex] for any point on the hyperbola. The derivation of the equation of a hyperbola is based on applying the distance formula, but is again beyond the scope of this text. The standard form of an equation of a hyperbola centered at the origin with vertices [latex]\left(\pm a,0\right)[/latex] and co-vertices [latex]\left(0\pm b\right)[/latex] is [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex].

A General Note: Standard Forms of the Equation of a Hyperbola with Center (0,0)

The standard form of the equation of a hyperbola with center [latex]\left(0,0\right)[/latex] and transverse axis on the x-axis is

[latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex]

where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(\pm a,0\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(0,\pm b\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(\pm c,0\right)[/latex]
  • the equations of the asymptotes are [latex]y=\pm \dfrac{b}{a}x[/latex]

The standard form of the equation of a hyperbola with center [latex]\left(0,0\right)[/latex] and transverse axis on the y-axis is

[latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex]

where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(0,\pm a\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(\pm b,0\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(0,\pm c\right)[/latex]
  • the equations of the asymptotes are [latex]y=\pm \dfrac{a}{b}x[/latex]

Note that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

(a) Horizontal hyperbola with center [latex]\left(0,0\right)[/latex] (b) Vertical hyperbola with center [latex]\left(0,0\right)[/latex]

How To: Given the equation of a hyperbola in standard form, locate its vertices and foci.

  1. Determine whether the transverse axis lies on the x– or y-axis. Notice that [latex]{a}^{2}[/latex] is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.

a. If the equation has the form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex], then the transverse axis lies on the x-axis. The vertices are located at [latex]\left(\pm a,0\right)[/latex], and the foci are located at [latex]\left(\pm c,0\right)[/latex].

b. If the equation has the form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex], then the transverse axis lies on the y-axis. The vertices are located at [latex]\left(0,\pm a\right)[/latex], and the foci are located at [latex]\left(0,\pm c\right)[/latex].

  1. Solve for [latex]a[/latex] using the equation [latex]a=\sqrt{{a}^{2}}[/latex].
  2. Solve for [latex]c[/latex] using the equation [latex]c=\sqrt{{a}^{2}+{b}^{2}}[/latex].

Example: Locating a Hyperbola’s Vertices and Foci

Identify the vertices and foci of the hyperbola with equation [latex]\dfrac{{y}^{2}}{49}-\dfrac{{x}^{2}}{32}=1[/latex].

Show Solution

The equation has the form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex], so the transverse axis lies on the y-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set [latex]x=0[/latex], and solve for [latex]y[/latex].

[latex]\begin{align}1=\frac{{y}^{2}}{49}-\frac{{x}^{2}}{32} \\[1mm] 1=\frac{{y}^{2}}{49}-\frac{{0}^{2}}{32} \\[1mm] 1=\frac{{y}^{2}}{49} \\[1mm] {y}^{2}=49 \\[1mm] y=\pm \sqrt{49}=\pm 7 \end{align}[/latex]

The foci are located at [latex]\left(0,\pm c\right)[/latex]. Solving for [latex]c[/latex],

[latex]c=\sqrt{{a}^{2}+{b}^{2}}=\sqrt{49+32}=\sqrt{81}=9[/latex]

Therefore, the vertices are located at [latex]\left(0,\pm 7\right)[/latex], and the foci are located at [latex]\left(0,9\right)[/latex].

Try It

Identify the vertices and foci of the hyperbola with equation [latex]\dfrac{{x}^{2}}{9}-\dfrac{{y}^{2}}{25}=1[/latex].

Show Solution

Vertices: [latex]\left(\pm 3,0\right)[/latex]; Foci: [latex]\left(\pm \sqrt{34},0\right)[/latex]

Writing Equations of Hyperbolas in Standard Form

Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.

Hyperbolas Centered at the Origin

Reviewing the standard forms given for hyperbolas centered at [latex]\left(0,0\right)[/latex], we see that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. Note that this equation can also be rewritten as [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex]. This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.

How To: Given the vertices and foci of a hyperbola centered at [latex]\left(0,\text{0}\right)[/latex], write its equation in standard form.

  1. Determine whether the transverse axis lies on the x– or y-axis.
    1. If the given coordinates of the vertices and foci have the form [latex]\left(\pm a,0\right)[/latex] and [latex]\left(\pm c,0\right)[/latex], respectively, then the transverse axis is the x-axis. Use the standard form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex].
    2. If the given coordinates of the vertices and foci have the form [latex]\left(0,\pm a\right)[/latex] and [latex]\left(0,\pm c\right)[/latex], respectively, then the transverse axis is the y-axis. Use the standard form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex].
  2. Find [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex].
  3. Substitute the values for [latex]{a}^{2}[/latex] and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.

Example: Finding the Equation of a Hyperbola Centered at (0,0) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices [latex]\left(\pm 6,0\right)[/latex] and foci [latex]\left(\pm 2\sqrt{10},0\right)?[/latex]

Show Solution

The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex].

The vertices are [latex]\left(\pm 6,0\right)[/latex], so [latex]a=6[/latex] and [latex]{a}^{2}=36[/latex].

The foci are [latex]\left(\pm 2\sqrt{10},0\right)[/latex], so [latex]c=2\sqrt{10}[/latex] and [latex]{c}^{2}=40[/latex].

Solving for [latex]{b}^{2}[/latex], we have

[latex]\begin{align}&{b}^{2}={c}^{2}-{a}^{2} \\ &{b}^{2}=40 - 36 && \text{Substitute for }{c}^{2}\text{ and }{a}^{2}. \\ &{b}^{2}=4 && \text{Subtract}. \end{align}[/latex]

Finally, we substitute [latex]{a}^{2}=36[/latex] and [latex]{b}^{2}=4[/latex] into the standard form of the equation, [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex]. The equation of the hyperbola is [latex]\dfrac{{x}^{2}}{36}-\dfrac{{y}^{2}}{4}=1[/latex].

Try It

What is the standard form equation of the hyperbola that has vertices [latex]\left(0,\pm 2\right)[/latex] and foci [latex]\left(0,\pm 2\sqrt{5}\right)?[/latex]

Show Solution

[latex]\dfrac{{y}^{2}}{4}-\dfrac{{x}^{2}}{16}=1[/latex]

Hyperbolas Not Centered at the Origin

Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated [latex]h[/latex] units horizontally and [latex]k[/latex] units vertically, the center of the hyperbola will be [latex]\left(h,k\right)[/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[/latex] replaced by [latex]\left(x-h\right)[/latex] and [latex]y[/latex] replaced by [latex]\left(y-k\right)[/latex].

A General Note: Standard Forms of the Equation of a Hyperbola with Center (h, k)

The standard form of the equation of a hyperbola with center [latex]\left(h,k\right)[/latex] and transverse axis parallel to the x-axis is
[latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]
where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(h\pm a,k\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex]

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is [latex]2a[/latex] and its width is [latex]2b[/latex]. The slopes of the diagonals are [latex]\pm \frac{b}{a}[/latex], and each diagonal passes through the center [latex]\left(h,k\right)[/latex]. Using the point-slope formula, it is simple to show that the equations of the asymptotes are [latex]y=\pm \frac{b}{a}\left(x-h\right)+k[/latex].

The standard form of the equation of a hyperbola with center [latex]\left(h,k\right)[/latex] and transverse axis parallel to the y-axis is
[latex]\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}-\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex]
where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(h,k\pm a\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex]

Using the reasoning above, the equations of the asymptotes are [latex]y=\pm \frac{a}{b}\left(x-h\right)+k[/latex].

(a) Horizontal hyperbola with center [latex]\left(h,k\right)[/latex] (b) Vertical hyperbola with center [latex]\left(h,k\right)[/latex]

Like hyperbolas centered at the origin, hyperbolas centered at a point [latex]\left(h,k\right)[/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.

How To: Given the vertices and foci of a hyperbola centered at [latex]\left(h,k\right)[/latex], write its equation in standard form.

  1. Determine whether the transverse axis is parallel to the x– or y-axis.
    1. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form [latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex].
    2. If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form [latex]\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}-\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex].
  2. Identify the center of the hyperbola, [latex]\left(h,k\right)[/latex], using the midpoint formula and the given coordinates for the vertices.
  3. Find [latex]{a}^{2}[/latex] by solving for the length of the transverse axis, [latex]2a[/latex] , which is the distance between the given vertices.
  4. Find [latex]{c}^{2}[/latex] using [latex]h[/latex] and [latex]k[/latex] found in Step 2 along with the given coordinates for the foci.
  5. Solve for [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex].
  6. Substitute the values for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.

Example: Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices at [latex]\left(0,-2\right)[/latex] and [latex]\left(6,-2\right)[/latex] and foci at [latex]\left(-2,-2\right)[/latex] and [latex]\left(8,-2\right)?[/latex]

Show Solution

The y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x-axis. Thus, the equation of the hyperbola will have the form

 

[latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]

First, we identify the center, [latex]\left(h,k\right)[/latex]. The center is halfway between the vertices [latex]\left(0,-2\right)[/latex] and [latex]\left(6,-2\right)[/latex]. Applying the midpoint formula, we have

[latex]\left(h,k\right)=\left(\dfrac{0+6}{2},\dfrac{-2+\left(-2\right)}{2}\right)=\left(3,-2\right)[/latex]

Next, we find [latex]{a}^{2}[/latex]. The length of the transverse axis, [latex]2a[/latex], is bounded by the vertices. So, we can find [latex]{a}^{2}[/latex] by finding the distance between the x-coordinates of the vertices.

[latex]\begin{gathered}2a=|0 - 6| \\ 2a=6 \\ a=3 \\ {a}^{2}=9 \end{gathered}[/latex]

Now we need to find [latex]{c}^{2}[/latex]. The coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex]. So [latex]\left(h-c,k\right)=\left(-2,-2\right)[/latex] and [latex]\left(h+c,k\right)=\left(8,-2\right)[/latex]. We can use the x-coordinate from either of these points to solve for [latex]c[/latex]. Using the point [latex]\left(8,-2\right)[/latex], and substituting [latex]h=3[/latex],

[latex]\begin{gathered}h+c=8 \\ 3+c=8 \\ c=5 \\ {c}^{2}=25 \end{gathered}[/latex]

Next, solve for [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}:[/latex]

[latex]\begin{align}{b}^{2}&={c}^{2}-{a}^{2} \\ &=25 - 9 \\ &=16 \end{align}[/latex]

Finally, substitute the values found for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation.

[latex]\dfrac{{\left(x - 3\right)}^{2}}{9}-\dfrac{{\left(y+2\right)}^{2}}{16}=1[/latex]

Try It

What is the standard form equation of the hyperbola that has vertices [latex]\left(1,-2\right)[/latex] and [latex]\left(1,\text{8}\right)[/latex] and foci [latex]\left(1,-10\right)[/latex] and [latex]\left(1,16\right)?[/latex]

Show Solution

[latex]\dfrac{{\left(y - 3\right)}^{2}}{25}-\dfrac{{\left(x - 1\right)}^{2}}{144}=1[/latex]

Solving Applied Problems Involving Hyperbolas

As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. For example a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide!

description in caption

Cooling towers at the Drax power station in North Yorkshire, United Kingdom (credit: Les Haines, Flickr)

The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In Example 6 we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides.

Example: Solving Applied Problems Involving Hyperbolas

The design layout of a cooling tower is shown below. The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart.

Project design for a natural draft cooling tower

Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

Show Solution

We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin: [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex], where the branches of the hyperbola form the sides of the cooling tower. We must find the values of [latex]{a}^{2}[/latex] and [latex]{b}^{2}[/latex] to complete the model.

First, we find [latex]{a}^{2}[/latex]. Recall that the length of the transverse axis of a hyperbola is [latex]2a[/latex]. This length is represented by the distance where the sides are closest, which is given as [latex]\text{ }65.3\text{ }[/latex] meters. So, [latex]2a=60[/latex]. Therefore, [latex]a=30[/latex] and [latex]{a}^{2}=900[/latex].

To solve for [latex]{b}^{2}[/latex], we need to substitute for [latex]x[/latex] and [latex]y[/latex] in our equation using a known point. To do this, we can use the dimensions of the tower to find some point [latex]\left(x,y\right)[/latex] that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y-axis bisects the tower, our x-value can be represented by the radius of the top, or 36 meters. The y-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore,

[latex]\begin{align}\dfrac{{x}^{2}}{{a}^{2}}&-\dfrac{{y}^{2}}{{b}^{2}}=1 && \text{Standard form of horizontal hyperbola}. \\ {b}^{2}&=\frac{{y}^{2}}{\frac{{x}^{2}}{{a}^{2}}-1} && \text{Isolate }{b}^{2} \\ &=\frac{{\left(79.6\right)}^{2}}{\frac{{\left(36\right)}^{2}}{900}-1} && \text{Substitute for }{a}^{2},x,\text{ and }y \\ &\approx 14400.3636 && \text{Round to four decimal places} \end{align}[/latex]

The sides of the tower can be modeled by the hyperbolic equation

[latex]\dfrac{{x}^{2}}{900}-\dfrac{{y}^{2}}{14400.3636 }=1,\text{ or }\dfrac{{x}^{2}}{{30}^{2}}-\dfrac{{y}^{2}}{{120.0015}^{2} }=1[/latex]

Try It

A design for a cooling tower project is shown below. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.

Show Solution

The sides of the tower can be modeled by the hyperbolic equation. [latex]\dfrac{{x}^{2}}{400}-\dfrac{{y}^{2}}{3600}=1\text{ or }\dfrac{{x}^{2}}{{20}^{2}}-\dfrac{{y}^{2}}{{60}^{2}}=1[/latex].

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Sours: https://courses.lumenlearning.com/waymakercollegealgebra/chapter/equations-of-hyperbolas/
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Hyperbola

FAQs on Hyperbola

What is Hyperbola in Conic Section?

A hyperbola is the locus of a point whose difference of the distances from two fixed points is a constant value. The two fixed points are called the foci of the hyperbola, and the equation of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\). Here a is called the semi-major axis and b is called the semi-minor axis of the hyperbola.

What is the Equation of Hyperbola?

The equation of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\). Here 'a' is the sem-major axis, and 'b' is the semi-minor axis. There are two standard forms of equations of a hyperbola.

How to Find the General Equation of a Hyperbola?

The equation of the hyperbola can be derived from the basic definition of a hyperbola: A hyperbola is the locus of a point whose difference of the distances from two fixed points is a constant value. Let the fixed point be P(x, y), the foci are F and F'. Then the condition is PF - PF' = 2a. The difference is taken from the farther focus, and then the nearer focus. This on further substitutions and simplification we have the equation of the hyperbola as \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\).

What is a Rectangular Hyperbola?

The hyperbola having the major axis and the minor axis of equal length is called a rectangular hyperbola. Hence we have 2a = 2b, or a = b. The equation of the rectangular hyperbola is x2 - y2 = a2.

What is the Standard Equation of Hyperbola?

There are two standard equations of the Hyperbola. These equations are given as,

  • \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\), for an hyperbola having the transverse axis as the x-axis and the conjugate axis is the y-axis.
  • \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\), for an hyperbola having the transverse axis as the y-axis and its conjugate axis is the x-axis.

What is the Eccentricity of Hyperbola?

The eccentricity of the hyperbola is greater than 1. (e > 1). The eccentricity is the ratio of the distance of the focus from the center of the ellipse, and the distance of the vertex from the center of the ellipse. The distance of the focus is 'c' units, and the distance of the vertex is 'a' units, and hence the eccentricity is e = c/a. Also here we have c2 = a2 + b2.

What is the Foci of a Hyperbola?

The hyperbola has two foci on either side of its center, and on its transverse axis. The hyperbola \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) has two foci (c, 0), and (-c, 0).

What is the Conjugate Axis of a Hyperbola?

The axis line passing through the center of the hyperbola and perpendicular to its transverse axis is called the conjugate axis of the hyperbola. The conjugate axis of the hyperbola having the equation \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) is the y-axis.

What are Asymptotes of Hyperbola?

The asymptotes are the lines that are parallel to the hyperbola and are assumed to meet the hyperbola at infinity. The equation of asymptotes of the hyperbola are y = bx/a, and y = -bx/a. The equation of pair of asymptotes of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0\).

What are the Vertices of a Hyperbola?

The vertices of a hyperbola are the points where the hyperbola cuts its transverse axis. The hyperbola has only two vertices, and the vertices of the hyperbola \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) is (a, 0), and (-a, 0) respectively.

How to Find Transverse Axis of a Hyperbola?

The transverse axis of a hyperbola is a line passing through the center and the two foci of the hyperbola. A hyperbola with an equation \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) had the x-axis as its transverse axis.

Sours: https://www.cuemath.com/geometry/hyperbola/

Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.

Hyperbolas Centered at the Origin

Reviewing the standard forms given for hyperbolas centered at [latex]\left(0,0\right)[/latex], we see that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. Note that this equation can also be rewritten as [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex]. This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.

How To: Given the vertices and foci of a hyperbola centered at [latex]\left(0,\text{0}\right)[/latex], write its equation in standard form.

  1. Determine whether the transverse axis lies on the x– or y-axis.
    1. If the given coordinates of the vertices and foci have the form [latex]\left(\pm a,0\right)[/latex] and [latex]\left(\pm c,0\right)[/latex], respectively, then the transverse axis is the x-axis. Use the standard form [latex]\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1[/latex].
    2. If the given coordinates of the vertices and foci have the form [latex]\left(0,\pm a\right)[/latex] and [latex]\left(0,\pm c\right)[/latex], respectively, then the transverse axis is the y-axis. Use the standard form [latex]\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1[/latex].
  2. Find [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex].
  3. Substitute the values for [latex]{a}^{2}[/latex] and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.

Example 2: Finding the Equation of a Hyperbola Centered at (0,0) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices [latex]\left(\pm 6,0\right)[/latex] and foci [latex]\left(\pm 2\sqrt{10},0\right)?[/latex]

Solution

The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form [latex]\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1[/latex].

The vertices are [latex]\left(\pm 6,0\right)[/latex], so [latex]a=6[/latex] and [latex]{a}^{2}=36[/latex].

The foci are [latex]\left(\pm 2\sqrt{10},0\right)[/latex], so [latex]c=2\sqrt{10}[/latex] and [latex]{c}^{2}=40[/latex].

Solving for [latex]{b}^{2}[/latex], we have

[latex]\begin{array}{ll}{b}^{2}={c}^{2}-{a}^{2}\hfill & \hfill \\ {b}^{2}=40 - 36\hfill & \begin{array}{cccc}& & & \end{array}\text{Substitute for }{c}^{2}\text{ and }{a}^{2}.\hfill \\ {b}^{2}=4\hfill & \begin{array}{cccc}& & & \end{array}\text{Subtract}.\hfill \end{array}[/latex]

Finally, we substitute [latex]{a}^{2}=36[/latex] and [latex]{b}^{2}=4[/latex] into the standard form of the equation, [latex]\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1[/latex]. The equation of the hyperbola is [latex]\frac{{x}^{2}}{36}-\frac{{y}^{2}}{4}=1[/latex].

Figure 6

Try It 2

What is the standard form equation of the hyperbola that has vertices [latex]\left(0,\pm 2\right)[/latex] and foci [latex]\left(0,\pm 2\sqrt{5}\right)?[/latex]

Solution

Hyperbolas Not Centered at the Origin

Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated [latex]h[/latex] units horizontally and [latex]k[/latex] units vertically, the center of the hyperbola will be [latex]\left(h,k\right)[/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[/latex] replaced by [latex]\left(x-h\right)[/latex] and [latex]y[/latex] replaced by [latex]\left(y-k\right)[/latex].

A General Note: Standard Forms of the Equation of a Hyperbola with Center (h, k)

The standard form of the equation of a hyperbola with center [latex]\left(h,k\right)[/latex] and transverse axis parallel to the x-axis is

[latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]

where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(h\pm a,k\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex]

The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is [latex]2a[/latex] and its width is [latex]2b[/latex]. The slopes of the diagonals are [latex]\pm \frac{b}{a}[/latex], and each diagonal passes through the center [latex]\left(h,k\right)[/latex]. Using the point-slope formula, it is simple to show that the equations of the asymptotes are [latex]y=\pm \frac{b}{a}\left(x-h\right)+k[/latex].

The standard form of the equation of a hyperbola with center [latex]\left(h,k\right)[/latex] and transverse axis parallel to the y-axis is

[latex]\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex]

where

  • the length of the transverse axis is [latex]2a[/latex]
  • the coordinates of the vertices are [latex]\left(h,k\pm a\right)[/latex]
  • the length of the conjugate axis is [latex]2b[/latex]
  • the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)[/latex]
  • the distance between the foci is [latex]2c[/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]
  • the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex]

Using the reasoning above, the equations of the asymptotes are [latex]y=\pm \frac{a}{b}\left(x-h\right)+k[/latex].

Figure 7. (a) Horizontal hyperbola with center [latex]\left(h,k\right)[/latex] (b) Vertical hyperbola with center [latex]\left(h,k\right)[/latex]

Like hyperbolas centered at the origin, hyperbolas centered at a point [latex]\left(h,k\right)[/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[/latex]. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.

How To: Given the vertices and foci of a hyperbola centered at [latex]\left(h,k\right)[/latex], write its equation in standard form.

  1. Determine whether the transverse axis is parallel to the x– or y-axis.
    1. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form [latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex].
    2. If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form [latex]\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex].
  2. Identify the center of the hyperbola, [latex]\left(h,k\right)[/latex], using the midpoint formula and the given coordinates for the vertices.
  3. Find [latex]{a}^{2}[/latex] by solving for the length of the transverse axis, [latex]2a[/latex] , which is the distance between the given vertices.
  4. Find [latex]{c}^{2}[/latex] using [latex]h[/latex] and [latex]k[/latex] found in Step 2 along with the given coordinates for the foci.
  5. Solve for [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[/latex].
  6. Substitute the values for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation determined in Step 1.

Example 3: Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices at [latex]\left(0,-2\right)[/latex] and [latex]\left(6,-2\right)[/latex] and foci at [latex]\left(-2,-2\right)[/latex] and [latex]\left(8,-2\right)?[/latex]

Solution

The y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x-axis. Thus, the equation of the hyperbola will have the form

[latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex]

First, we identify the center, [latex]\left(h,k\right)[/latex]. The center is halfway between the vertices [latex]\left(0,-2\right)[/latex] and [latex]\left(6,-2\right)[/latex]. Applying the midpoint formula, we have

[latex]\left(h,k\right)=\left(\frac{0+6}{2},\frac{-2+\left(-2\right)}{2}\right)=\left(3,-2\right)[/latex]

Next, we find [latex]{a}^{2}[/latex]. The length of the transverse axis, [latex]2a[/latex], is bounded by the vertices. So, we can find [latex]{a}^{2}[/latex] by finding the distance between the x-coordinates of the vertices.

[latex]\begin{array}{l}2a=|0 - 6|\hfill \\ 2a=6\hfill \\ \text{ }a=3\hfill \\ {a}^{2}=9\hfill \end{array}[/latex]

Now we need to find [latex]{c}^{2}[/latex]. The coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex]. So [latex]\left(h-c,k\right)=\left(-2,-2\right)[/latex] and [latex]\left(h+c,k\right)=\left(8,-2\right)[/latex]. We can use the x-coordinate from either of these points to solve for [latex]c[/latex]. Using the point [latex]\left(8,-2\right)[/latex], and substituting [latex]h=3[/latex],

[latex]\begin{array}{l}h+c=8\hfill \\ 3+c=8\hfill \\ \text{ }c=5\hfill \\ \text{ }{c}^{2}=25\hfill \end{array}[/latex]

Next, solve for [latex]{b}^{2}[/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}:[/latex]

[latex]\begin{array}{l}{b}^{2}={c}^{2}-{a}^{2}\hfill \\ \text{ }=25 - 9\hfill \\ \text{ }=16\hfill \end{array}[/latex]

Finally, substitute the values found for [latex]h,k,{a}^{2}[/latex], and [latex]{b}^{2}[/latex] into the standard form of the equation.

[latex]\frac{{\left(x - 3\right)}^{2}}{9}-\frac{{\left(y+2\right)}^{2}}{16}=1[/latex]

Try It 3

What is the standard form equation of the hyperbola that has vertices [latex]\left(1,-2\right)[/latex] and [latex]\left(1,\text{8}\right)[/latex] and foci [latex]\left(1,-10\right)[/latex] and [latex]\left(1,16\right)?[/latex]

Solution

Sours: https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/writing-equations-of-hyperbolas-in-standard-form/

Form hyperbola standard

22.5: Hyperbolas

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Learning Objectives

By the end of this section, you will be able to:

  • Graph a hyperbola with center at \((0,0)\)
  • Graph a hyperbola with center at \((h,k)\)
  • Identify conic sections by their equations

Before you get started, take this readiness quiz.

  1. Solve: \(x^{2}=12\).
    If you missed this problem, review Example 9.1.
  2. Expand: \((x−4)^{2}\).
    If you missed this problem, review Example 5.32.
  3. Graph \(y=-\frac{2}{3} x\).
    If you missed this problem, review Example 3.4.

Graph a Hyperbola with Center at \((0,0)\)

The last conic section we will look at is called a hyperbola. We will see that the equation of a hyperbola looks the same as the equation of an ellipse, except it is a difference rather than a sum. While the equations of an ellipse and a hyperbola are very similar, their graphs are very different.

We define a hyperbola as all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a focus of the hyperbola.

Definition \(\PageIndex{1}\)

A hyperbola is all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a focus of the hyperbola.

The figure shows a double napped right circular cone sliced by a plane that is parallel to the vertical axis of the cone forming a hyperbola. The figure is labeled ‘hyperbola’.

The line through the foci, is called the transverse axis. The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola. The midpoint of the segment joining the foci is called the center of the hyperbola. The line perpendicular to the transverse axis that passes through the center is called the conjugate axis. Each piece of the graph is called a branch of the hyperbola.

The figure shows two graphs of a hyperbola. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci are shown with points that lie on the transverse axis, which is the x-axis. The branches pass through the vertices and open left and right. The y-axis is the conjugate axis. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci lie are shown with points that lie on the transverse axis, which is the y-axis. The branches pass through the vertices and open up and down. The x-axis is the conjugate axis.

Again our goal is to connect the geometry of a conic with algebra. Placing the hyperbola on a rectangular coordinate system gives us that opportunity. In the figure, we placed the hyperbola so the foci \(((−c,0),(c,0))\) are on the \(x\)-axis and the center is the origin.

The figure shows the graph of a hyperbola. The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The foci (negative c, 0) and (c, 0) are marked with a point and lie on the x-axis. The vertices are marked with a point and lie on the x-axis. The branches pass through the vertices and open left and right. The distance from (negative c, 0) to a point on the branch (x, y) is marked d sub 1. The distance from (x, y) on the branch to (c, 0) is marked d sub 2.

The definition states the difference of the distance from the foci to a point \((x,y)\) is constant. So \(|d_{1}−d_{2}|\) is a constant that we will call \(2a\) so \(|d_{1}-d_{2} |=2 a\). We will use the distance formula to lead us to an algebraic formula for an ellipse.

\(\left|d_{1} - d_{2}\right| =2 a\)

Use the distance formula to find \(d_{1}, d_{2}\)

\(\left|\sqrt{(x-(-c))^{2}+(y-0)^{2}}-\sqrt{(x-c)^{2}+(y-0)^{2}}\right|=2 a\)

Eliminate the radicals. To simplify the equation of the ellipse, we let \(c^{2}-a^{2}=b^{2}\).

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{c^{2}-a^{2}}=1\)

So, the equation of a hyperbola centered at the origin in standard form is:

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

To graph the hyperbola, it will be helpful to know about the intercepts. We will find the \(x\)-intercepts and \(y\)-intercepts using the formula.

\(x\)-intercepts

Let \(y=0\).

\(\begin{aligned} \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} &=1 \\ \frac{x^{2}}{a^{2}}-\frac{0^{2}}{b^{2}} &=1 \\ \frac{x^{2}}{a^{2}} &=1 \\ x^{2} &=a^{2} \\ x &=\pm a \end{aligned}\)

The \(x\)-intercepts are \((a,0)\) and \((−a,0)\).

\(y\)-intercepts

Let \(x=0\).

\(\begin{aligned} \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} &=1 \\ \frac{0^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} &=1 \\-\frac{y^{2}}{b^{2}} &=1 \\ y^{2} &=-b^{2} \\ y &=\pm \sqrt{-b^{2}} \end{aligned}\)

There are no \(y\)-intercepts.

The \(a, b\) values in the equation also help us find the asymptotes of the hyperbola. The asymptotes are intersecting straight lines that the branches of the graph approach but never intersect as the \(x, y\) values get larger and larger.

To find the asymptotes, we sketch a rectangle whose sides intersect the x-axis at the vertices \((−a,0),(a,0)\), and intersect the \(y\)-axis at \((0,−b), (0,b)\). The lines containing the diagonals of this rectangle are the asymptotes of the hyperbola. The rectangle and asymptotes are not part of the hyperbola, but they help us graph the hyperbola.

The figure shows the graph of a hyperbola. The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (negative a, 0) and (a, 0) and are marked with a point and lie on the x-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the x-axis at the vertices (negative a, 0) and (a, 0) and intersect the y-axis at (0, b) and (0, negative b). The asymptotes are given by y is equal to b divided by a times x and y is equal to negative b divided by a times x and are drawn as the diagonals of the central rectangle. The branches of the hyperbola pass through the vertices, open left and right, and approach the asymptotes.

The asymptotes pass through the origin and we can evaluate their slope using the rectangle we sketched. They have equations \(y=\frac{b}{a} x\) and \(y=-\frac{b}{a} x\).

There are two equations for hyperbolas, depending whether the transverse axis is vertical or horizontal. We can tell whether the transverse axis is horizontal by looking at the equation. When the equation is in standard form, if the \(x^{2}\)-term is positive, the transverse axis is horizontal. When the equation is in standard form, if the \(y^{2}\)-term is positive, the transverse axis is vertical.

The second equations could be derived similarly to what we have done. We will summarize the results here.

Definition \(\PageIndex{2}\)

Standard Form of the Equation a Hyperbola with Center \((0,0)\)

The standard form of the equation of a hyperbola with center \((0,0)\), is

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \quad\) or \(\quad \frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\)

The figure shows the graph of two hyperbolas. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (negative a, 0) and (a, 0) and are marked with a point and lie on the x-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the x-axis at the vertices (negative a, 0) and (a, 0) and intersect the y-axis at (0, b) and (0, negative b). The asymptotes are given by y is equal to b divided by a times x and y is equal to negative b divided by a times x and are drawn as the diagonals of the central rectangle. The branches of the hyperbola pass through the vertices, open left and right, and approach the asymptotes. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (0, a) and (0, negative a) and are marked with a point and lie on the y-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the y-axis at the vertices (0, a) and (0, negative a) and intersect the y-axis at (negative b, 0) and (b, 0). The branches of the hyperbola pass through the vertices, open up and down, and approach the asymptotes.

Notice that, unlike the equation of an ellipse, the denominator of \(x^{2}\) is not always \(a^{2}\) and the denominator of \(y^{2}\) is not always \(b^{2}\).

Notice that when the \(x^{2}\)-term is positive, the transverse axis is on the \(x\)-axis. When the \(y^{2}\)-term is positive, the transverse axis is on the \(y\)-axis.

Standard Forms of the Equation a Hyperbola with Center \((0,0)\)

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)\(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\)
OrientationTransverse axis on the \(x\)-axis.
Opens left and right
Transverse axis on the \(y\)-axis.
Opens up and down
Vertices\((-a, 0),(a, 0)\)\((0,-a),(0, a)\)
\(x\)-intercepts\((-a, 0),(a, 0)\)none
\(y\)-interceptsnone\((0,-a),(0, a)\)
RectangleUse \(( \pm a, 0)(0, \pm b)\)Use \((0, \pm a)( \pm b, 0)\)
Asymptotes\(y=\frac{b}{a} x, y=-\frac{b}{a} x\)\(y=\frac{a}{b} x, y=-\frac{a}{b} x\)

We will use these properties to graph hyperbolas.

Example \(\PageIndex{1}\) How to Graph a Hyperbola with Center \((0,0)\)

Graph \(\frac{x^{2}}{25}-\frac{y^{2}}{4}=1\).

Solution:

Step 1: Write the equation in standard form.The equation is in standard form.\(\frac{x^{2}}{25}-\frac{y^{2}}{4}=1\)
Step 2: Determine whether the transverse axis is horizontal or vertical.Since the \(x^{2}\)-term is positive, the transverse axis is horizontal.The transverse axis is horizontal.
Step 3: Find the vertices.Since \(a^{2}=25\) then \(a=\pm 5\). The vertices are on the \(x\)-axis.\((-5,0),(5,0)\)
Step 4: Sketch the rectangle centered at the origin intersection one axis at \(\pm a\) and the other at \(\pm b\).

Since \(a=\pm 5\), the rectangle will intersect the \(x\)-axis at the vertices.

Since \(b=\pm 2\), the rectangle will intersect the \(y\)-axis at \((0,-2)\) and \((0,2)\).

Screenshot (148).png

Step 5: Sketch the asymptotes--the lines through the diagonals of the rectangle.

The asymptotes have the equations \(y=\frac{5}{2} x, y=-\frac{5}{2} x\).Screenshot (149).png
Step 6: Draw the two branches of the hyperbola.Start at each vertex and use the asymptotes as a guide.Screenshot (150).png

Exercise \(\PageIndex{1}\)

Graph \(\frac{x^{2}}{16}-\frac{y^{2}}{4}=1\).

Answer
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus one-half times x, and branches that pass through the vertices (plus or minus 4, 0) and open left and right.

Exercise \(\PageIndex{2}\)

Graph \(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\).

Answer
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus four-thirds times x, and branches that pass through the vertices (plus or minus 3, 0) and open left and right.

We summarize the steps for reference.

Graph a Hyperbola Centered at \((0,0)\)

  1. Write the equation in standard form.
  2. Determine whether the transverse axis is horizontal or vertical.
  3. Find the vertices.
  4. Sketch the rectangle centered at the origin intersecting one axis at \(±a\) and the other at \(±b\).
  5. Sketch the asymptotes—the lines through the diagonals of the rectangle.
  6. Draw the two branches of the hyperbola.

Sometimes the equation for a hyperbola needs to be first placed in standard form before we graph it.

Example \(\PageIndex{2}\)

Graph \(4 y^{2}-16 x^{2}=64\).

Solution:

\(4 y^{2}-16 x^{2}=64\)
To write the equation in standard form, divide each term by \(64\) to make the equation equal to \(1\).\(\frac{4 y^{2}}{64}-\frac{16 x^{2}}{64}=\frac{64}{64}\)
Simplify.\(\frac{y^{2}}{16}-\frac{x^{2}}{4}=1\)
Since the \(y^{2}\)-term is positive, the transverse axis is vertical. Since \(a^{2}=16\) then \(a=\pm 4\).
The vertices are on the \(y\)-axis, \((0,-a),(0, a)\). Since \(b^{2}=4\) then \(b=\pm 2\).\((0,-4),(0,4)\)
Sketch the rectangle intersecting the \(x\)-axis at \((-2,0),(2,0)\) and the \(y\)-axis at the vertices. Sketch the asymptotes through the diagonals of the rectangle. Draw the two branches of the hyperbola..

Exercise \(\PageIndex{3}\)

Graph \(4 y^{2}-25 x^{2}=100\).

Answer
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus five-halves times x, and branches that pass through the vertices (0, plus or minus 5) and open up and down.

Exercise \(\PageIndex{4}\)

Graph \(25 y^{2}-9 x^{2}=225\).

Answer
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with asymptotes y is equal to plus or minus three-fifths times x, and branches that pass through the vertices (0, plus or minus 3) and open up and down.

Graph a Hyperbola with Center at \((h,k)\)

Hyperbolas are not always centered at the origin. When a hyperbola is centered at \((h,k)\) the equations changes a bit as reflected in the table.

Standard Forms of the Equation a Hyperbola with Center \((h,k)\)

\(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\)\(\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1\)
OrientationTransverse axis is horizontal. Opens left and rightTransverse axis is vertical. Opens up and down
Center\((h,k)\)\((h,k)\)
Vertices\(a\) units to the left and right of the center\(a\) units above and below the center
RectangleUse \(a\) units left/right of center \(b\) units above/below the centerUse \(a\) units above/below the center \(b\) units left/right of center

Example \(\PageIndex{3}\) How to Graph a Hyperbola with Center \((h,k)\)

Graph \(\frac{(x-1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1\)

Solution:

Step 1: Write the equation in standard form.The equation is in standard form.\(\frac{(x-1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1\)
Step 2: Determine whether the transverse axis is horizontal or vertical.Since the \(x^{2}\)-term is positive, the hyperbola opens left and right.The transverse axis is horizontal. The hyperbola opens left and right.
Step 3: Find the center and \(a, b\).\(h=1\) and \(k=2\)
\(a^{2}=9\)
\(b^{2}=16\)

\(\begin{array} {c} \frac{\left(\stackrel{\color{red}{x-h}}{\color{black}{x-1}} \right)^{2}}{9} - \frac{\left(\stackrel{\color{red}{y-k}}{\color{black}{y-2}} \right)^{2}}{16} = 1 \end{array}\)

Center: \((1,2)\)

\(a=3\)

\(b=4\)

Step 4: Sketch the rectangle centered at \((h,k)\) using \(a,b\).

Mark the center, \((1,2)\).

Sketch the rectangle that goes through the points \(3\) units to the left/right of the center and \(4\) units above and below the center.

Screenshot (151).png
Step 5: Sketch the asymptotes--the lines through the diagonals of the rectangle. Mark the vertices.Sketch the diagonals. Mark the vertices, which are on the rectangle \(3\) units to the left and right of the center.Screenshot (152).png
Step 6: Draw the two branches of the hyperbola.Start at each vertex and use the asymptotes as a guide.Screenshot (153).png

Exercise \(\PageIndex{5}\)

Graph \(\frac{(x-3)^{2}}{25}-\frac{(y-1)^{2}}{9}=1\).

Answer
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with an asymptote that passes through (negative 2, negative 2) and (8, 4) and an asymptote that passes through (negative 2, 4) and (8, negative 2), and branches that pass through the vertices (negative 2, 2) and (8, 2) and opens left and right.

Exercise \(\PageIndex{6}\)

Graph \(\frac{(x-2)^{2}}{4}-\frac{(y-2)^{2}}{9}=1\).

Answer
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (2, 2), an asymptote that passes through (0, negative 1) and (4, 5) and an asymptote that passes through (0, 5) and (4, negative 1), and branches that pass through the vertices (0, 2) and (4, 2) and opens left and right.

We summarize the steps for easy reference.

Graph a Hyperbola Centered at \((h,k)\)

  1. Write the equation in standard form.
  2. Determine whether the transverse axis is horizontal or vertical.
  3. Find the center and \(a,b\).
  4. Sketch the rectangle centered at \((h,k)\) using \(a,b\).
  5. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices.
  6. Draw the two branches of the hyperbola.

Be careful as you identify the center. The standard equation has \(x−h\) and \(y−k\) with the center as \((h,k)\).

Example \(\PageIndex{4}\)

Graph \(\frac{(y+2)^{2}}{9}-\frac{(x+1)^{2}}{4}=1\).

Solution:

.
Since the \(y^{2}\)-term is positive, the hyperbola opens up and down..
Find the center, \((h,k)\).Center: \((-1,-2)\)
Find \(a,b\).\(a=3 b=2\)
Sketch the rectangle that goes through the points \(3\) units above and below the center and
\(2\) units to the left/right of the center.
Sketch the asymptotes—the lines through the diagonals of the rectangle.
Mark the vertices.
Graph the branches.
.

Exercise \(\PageIndex{7}\)

Graph \(\frac{(y+3)^{2}}{16}-\frac{(x+2)^{2}}{9}=1\).

Answer
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with a center at (negative 2, negative 3), an asymptote that passes through (negative 5, negative 7) and (1, 1) and an asymptote that passes through (negative 5, 1) and (1, 7), and branches that pass through the vertices (negative 2, 1) and (negative 2, negative 7) and opens up and down.

Exercise \(\PageIndex{8}\)

Graph \(\frac{(y+2)^{2}}{9}-\frac{(x+2)^{2}}{9}=1\).

Answer
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with a center at (negative 2, negative 2), an asymptote that passes through (negative 5, negative 5) and (1, 1) and an asymptote that passes through (negative 5, 1) and (1, negative 5), and branches that pass through the vertices (negative 2, 1) and (negative 2, negative 5) and opens up and down.

Again, sometimes we have to put the equation in standard form as our first step.

Example \(\PageIndex{5}\)

Write the equation in standard form and graph \(4 x^{2}-9 y^{2}-24 x-36 y-36=0\).

Solution:

Exercise \(\PageIndex{9}\)

  1. Write the equation in standard form and
  2. Graph \(9 x^{2}-16 y^{2}+18 x+64 y-199=0\).
Answer
  1. \(\frac{(x+1)^{2}}{16}-\frac{(y-2)^{2}}{9}=1\)
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (negative 1, 2), an asymptote that passes through (negative 5, 5) and (3, negative 1) and an asymptote that passes through (3, 5) and (negative 5, negative 1), and branches that pass through the vertices (negative 5, 2) and (3, 2) and opens left and right.

Exercise \(\PageIndex{10}\)

  1. Write the equation in standard form and
  2. Graph \(16 x^{2}-25 y^{2}+96 x-50 y-281=0\).
Answer
  1. \(\frac{(x+3)^{2}}{25}-\frac{(y+1)^{2}}{16}=1\)
The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals, with the center (negative 3, negative 1), an asymptote that passes through (negative 8, negative 5) and (2, 3) and an asymptote that passes through (negative 8, 3) and (2, negative 5), and branches that pass through the vertices (negative 8, negative 1) and (2, negative 1) and opens left and right.

Identify Conic Sections by Their Equations

Now that we have completed our study of the conic sections, we will take a look at the different equations and recognize some ways to identify a conic by its equation. When we are given an equation to graph, it is helpful to identify the conic so we know what next steps to take.

To identify a conic from its equation, it is easier if we put the variable terms on one side of the equation and the constants on the other.

ConicCharacteristics of \(x^{2}\)- and \(y^{2}\)-termsExample
ParabolaEither \(x^{2}\) OR \(y^{2}\). Only one variable is squared.\(x=3 y^{2}-2 y+1\)
Circle\(x^{2}\)- and \(y^{2}\)- terms have the same coefficients.\(x^{2}+y^{2}=49\)
Ellipse\(x^{2}\)- and \(y^{2}\)- terms have the same sign, different coefficients.\(4 x^{2}+25 y^{2}=100\)
Hyperbola\(x^{2}\)- and \(y^{2}\)- terms have the different signs, different coefficients.\(25 y^{2}-4 x^{2}=100\)

Example \(\PageIndex{6}\)

Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

  1. \(9 x^{2}+4 y^{2}+56 y+160=0\)
  2. \(9 x^{2}-16 y^{2}+18 x+64 y-199=0\)
  3. \(x^{2}+y^{2}-6 x-8 y=0\)
  4. \(y=-2 x^{2}-4 x-5\)

Solution:

a. The \(x^{2}\)- and \(y^{2}\)-terms have the same sign and different coefficients.

\(9 x^{2}+4 y^{2}+56 y+160=0\)

Ellipse

b. The \(x^{2}\)- and \(y^{2}\)-terms have different signs and different coefficients.

\(9 x^{2}-16 y^{2}+18 x+64 y-199=0\)

Hyperbola

c. The \(x^{2}\)- and \(y^{2}\)-terms have the same coefficients.

\(x^{2}+y^{2}-6 x-8 y=0\)

Circle

d. Only one variable, \(x\), is squared.

\(y=-2 x^{2}-4 x-5\)

Parabola

Exercise \(\PageIndex{11}\)

Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

  1. \(x^{2}+y^{2}-8 x-6 y=0\)
  2. \(4 x^{2}+25 y^{2}=100\)
  3. \(y=6 x^{2}+2 x-1\)
  4. \(16 y^{2}-9 x^{2}=144\)
Answer
  1. Circle
  2. Ellipse
  3. Parabola
  4. Hyperbola

Exercise \(\PageIndex{12}\)

Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

  1. \(16 x^{2}+9 y^{2}=144\)
  2. \(y=2 x^{2}+4 x+6\)
  3. \(x^{2}+y^{2}+2 x+6 y+9=0\)
  4. \(4 x^{2}-16 y^{2}=64\)
Answer
  1. Ellipse
  2. Parabola
  3. Circle
  4. Hyperbola

Access these online resources for additional instructions and practice with hyperbolas.

  • Graph a Hyperbola with Center at the Origin
  • Graph a Hyperbola with Center not at the Origin
  • Graph a Hyperbola in General Form
  • Identifying Conic Sections in General Form

Key Concepts

  • Hyperbola: A hyperbola is all points in a plane where the difference of their distances from two fixed points is constant.
The figure shows a double napped right circular cone sliced by a plane that is parallel to the vertical axis of the cone forming a hyperbola. The figure is labeled ‘hyperbola’.
  • Each of the fixed points is called a focus of the hyperbola.
    The line through the foci, is called the transverse axis.
    The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola.
    The midpoint of the segment joining the foci is called the center of the hyperbola.
    The line perpendicular to the transverse axis that passes through the center is called the conjugate axis.
    Each piece of the graph is called a branch of the hyperbola.
    The figure shows two graphs of a hyperbola. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci are shown with points that lie on the transverse axis, which is the x-axis. The branches pass through the vertices and open left and right. The y-axis is the conjugate axis. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci lie are shown with points that lie on the transverse axis, which is the y-axis. The branches pass through the vertices and open up and down. The x-axis is the conjugate axis.

Figure 11.4.2

Standard Forms of the Equation a Hyperbola with Center \((0,0)\)

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)\(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\)
OrientationTransverse axis on the \(x\)-axis.
Opens left and right
Transverse axis on the \(y\)-axis.
Opens up and down
Vertices\((-a, 0),(a, 0)\)\((0,-a),(0, a)\)
\(x\)-intercepts\((-a, 0),(a, 0)\)none
\(y\)-interceptsnone\((0,-a),(0, a)\)
RectangleUse \(( \pm a, 0)(0, \pm b)\)Use \((0, \pm a)( \pm b, 0)\)
Asymptotes\(y=\frac{b}{a} x, y=-\frac{b}{a} x\)\(y=\frac{a}{b} x, y=-\frac{a}{b} x\)
  • How to graph a hyperbola centered at \((0,0)\).
    1. Write the equation in standard form.
    2. Determine whether the transverse axis is horizontal or vertical.
    3. Find the vertices.
    4. Sketch the rectangle centered at the origin intersecting one axis at \(±a\) and the other at \(±b\).
    5. Sketch the asymptotes—the lines through the diagonals of the rectangle.
    6. Draw the two branches of the hyperbola.

Standard Forms of the Equation a Hyperbola with Center \((h,k)\)

\(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\)\(\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1\)
OrientationTransverse axis is horizontal. Opens left and rightTransverse axis is vertical. Opens up and down
Center\((h,k)\)\((h,k)\)
Vertices\(a\) units to the left and right of the center\(a\) units above and below the center
RectangleUse \(a\) units left/right of center \(b\) units above/below the centerUse \(a\) units above/below the center \(b\) units left/right of center
  • How to graph a hyperbola centered at \((h,k)\).
    1. Write the equation in standard form.
    2. Determine whether the transverse axis is horizontal or vertical.
    3. Find the center and \(a,b\).
    4. Sketch the rectangle centered at \((h,k)\) using \(a,b\).
    5. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices.
    6. Draw the two branches of the hyperbola.
ConicCharacteristics of \(x^{2}\)- and \(y^{2}\)-termsExample
ParabolaEither \(x^{2}\) OR \(y^{2}\). Only one variable is squared.\(x=3 y^{2}-2 y+1\)
Circle\(x^{2}\)- and \(y^{2}\)- terms have the same coefficients.\(x^{2}+y^{2}=49\)
Ellipse\(x^{2}\)- and \(y^{2}\)- terms have the same sign, different coefficients.\(4 x^{2}+25 y^{2}=100\)
Hyperbola\(x^{2}\)- and \(y^{2}\)- terms have the different signs, different coefficients.\(25 y^{2}-4 x^{2}=100\)

Glossary

hyperbola
A hyperbola is defined as all points in a plane where the difference of their distances from two fixed points is constant.
Sours: https://math.libretexts.org/Courses/Las_Positas_College/Foundational_Mathematics/22%3A_Conics/22.05%3A_Hyperbolas
How to Write a Hyperbola in Standard Form

The Hyperbola in Standard Form

A hyperbolaThe set of points in a plane whose distances from two fixed points, called foci, has an absolute difference that is equal to a positive constant. is the set of points in a plane whose distances from two fixed points, called foci, has an absolute difference that is equal to a positive constant. In other words, if points and are the foci and is some given positive constant then is a point on the hyperbola if as pictured below:

In addition, a hyperbola is formed by the intersection of a cone with an oblique plane that intersects the base. It consists of two separate curves, called branchesThe two separate curves of a hyperbola.. Points on the separate branches of the graph where the distance is at a minimum are called vertices.Points on the separate branches of a hyperbola where the distance is a minimum. The midpoint between a hyperbola’s vertices is its center. Unlike a parabola, a hyperbola is asymptotic to certain lines drawn through the center. In this section, we will focus on graphing hyperbolas that open left and right or upward and downward.

The asymptotes are drawn dashed as they are not part of the graph; they simply indicate the end behavior of the graph. The equation of a hyperbola opening left and right in standard formThe equation of a hyperbola written in the form The center is , a defines the transverse axis, and b defines the conjugate axis. follows:

Here the center is and the vertices are The equation of a hyperbola opening upward and downward in standard formThe equation of a hyperbola written in the form The center is , b defines the transverse axis, and a defines the conjugate axis. follows:

Here the center is and the vertices are

The asymptotes are essential for determining the shape of any hyperbola. Given standard form, the asymptotes are lines passing through the center with slope To easily sketch the asymptotes we make use of two special line segments through the center using a and b. Given any hyperbola, the transverse axisThe line segment formed by the vertices of a hyperbola. is the line segment formed by its vertices. The conjugate axisA line segment through the center of a hyperbola that is perpendicular to the transverse axis. is the line segment through the center perpendicular to the transverse axis as pictured below:

The rectangle defined by the transverse and conjugate axes is called the fundamental rectangleThe rectangle formed using the endpoints of a hyperbolas, transverse and conjugate axes.. The lines through the corners of this rectangle have slopes These lines are the asymptotes that define the shape of the hyperbola. Therefore, given standard form, many of the properties of a hyperbola are apparent.

Equation

Center

a

b

Opens

Left and right

Upward and downward

Upward and downward

Left and right

The graph of a hyperbola is completely determined by its center, vertices, and asymptotes.

Example 1

Graph:

Solution:

In this case, the expression involving x has a positive leading coefficient; therefore, the hyperbola opens left and right. Here and From the center , mark points 3 units left and right as well as 2 units up and down. Connect these points with a rectangle as follows:

The lines through the corners of this rectangle define the asymptotes.

Use these dashed lines as a guide to graph the hyperbola opening left and right passing through the vertices.

Answer:

Example 2

Graph:

Solution:

In this case, the expression involving y has a positive leading coefficient; therefore, the hyperbola opens upward and downward. Here and From the center mark points 6 units left and right as well as 2 units up and down. Connect these points with a rectangle. The lines through the corners of this rectangle define the asymptotes.

Use these dashed lines as a guide to graph the hyperbola opening upward and downward passing through the vertices.

Answer:

Note: When given a hyperbola opening upward and downward, as in the previous example, it is a common error to interchange the values for the center, h and k. This is the case because the quantity involving the variable y usually appears first in standard form. Take care to ensure that the y-value of the center comes from the quantity involving the variable y and that the x-value of the center is obtained from the quantity involving the variable x.

As with any graph, we are interested in finding the x- and y-intercepts.

Example 3

Find the intercepts:

Solution:

To find the x-intercepts set and solve for x.

Therefore there is only one x-intercept, To find the y-intercept set and solve for y.

Therefore there are two y-intercepts, and Take a moment to compare these to the sketch of the graph in the previous example.

Answer: x-intercept: ; y-intercepts: and

Consider the hyperbola centered at the origin,

Standard form requires one side to be equal to 1. In this case, we can obtain standard form by dividing both sides by 45.

This can be written as follows:

In this form, it is clear that the center is , , and The graph follows.

The Hyperbola in General Form

We have seen that the graph of a hyperbola is completely determined by its center, vertices, and asymptotes; which can be read from its equation in standard form. However, the equation is not always given in standard form. The equation of a hyperbola in general formThe equation of a hyperbola written in the form or where follows:

where The steps for graphing a hyperbola given its equation in general form are outlined in the following example.

Example 4

Graph:

Solution:

Begin by rewriting the equation in standard form.

  • Step 1: Group the terms with the same variables and move the constant to the right side. Factor so that the leading coefficient of each grouping is 1.

  • Step 2: Complete the square for each grouping. In this case, for the terms involving x use and for the terms involving y use The factor in front of each grouping affects the value used to balance the equation on the right,

    Because of the distributive property, adding 16 inside of the first grouping is equivalent to adding Similarly, adding 9 inside of the second grouping is equivalent to adding Now factor and then divide to obtain 1 on the right side.

  • Step 3: Determine the center, a, and b, and then use this information to sketch the graph. In this case, the center is , , and Because the leading coefficient of the expression involving x is positive and the coefficient of the expression involving y is negative, we graph a hyperbola opening left and right.

Answer:

Identifying the Conic Sections

In this section, the challenge is to identify a conic section given its equation in general form. To distinguish between the conic sections, use the exponents and coefficients. If the equation is quadratic in only one variable and linear in the other, then its graph will be a parabola.

Parabola:

Parabola:

If the equation is quadratic in both variables, where the coefficients of the squared terms are the same, then its graph will be a circle.

Circle:

If the equation is quadratic in both variables where the coefficients of the squared terms are different but have the same sign, then its graph will be an ellipse.

Ellipse: and

If the equation is quadratic in both variables where the coefficients of the squared terms have different signs, then its graph will be a hyperbola.

Hyperbola: and

Example 5

Identify the graph of each equation as a parabola, circle, ellipse, or hyperbola.

Solution:

  1. The equation is quadratic in both x and y where the leading coefficients for both variables is the same, 4.

    This is an equation of a circle centered at the origin with radius 1/2.

  2. The equation is quadratic in both x and y where the leading coefficients for both variables have different signs.

    This is an equation of a hyperbola opening left and right centered at the origin.

  3. The equation is quadratic in y only.

    This is an equation of a parabola opening right with vertex

Answer:

  1. Circle
  2. Hyperbola
  3. Parabola

Key Takeaways

  • The graph of a hyperbola is completely determined by its center, vertices, and asymptotes.
  • The center, vertices, and asymptotes are apparent if the equation of a hyperbola is given in standard form: or
  • To graph a hyperbola, mark points a units left and right from the center and points b units up and down from the center. Use these points to draw the fundamental rectangle; the lines through the corners of this rectangle are the asymptotes. If the coefficient of is positive, draw the branches of the hyperbola opening left and right through the points determined by a. If the coefficient of is positive, draw the branches of the hyperbola opening up and down through the points determined by b.
  • The orientation of the transverse axis depends the coefficient of and
  • If the equation of a hyperbola is given in general form or where , group the terms with the same variables, and complete the square for both groupings to obtain standard form.
  • We recognize the equation of a hyperbola if it is quadratic in both x and y where the coefficients of the squared terms are opposite in sign.

Topic Exercises

    Part A: The Hyperbola in Standard Form

      Given the equation of a hyperbola in standard form, determine its center, which way the graph opens, and the vertices.

      Determine the standard form for the equation of a hyperbola given the following information.

    1. Center , , , opens left and right.

    2. Center , , , opens up and down.

    3. Center , , , opens up and down.

    4. Center , , , opens left and right.

    5. Center , , ,opens up and down.

    6. Center , , , opens left and right.

      Graph.

      Find the x- and y-intercepts.

    1. Find the equation of the hyperbola with vertices and a conjugate axis that measures 12 units.

    2. Find the equation of the hyperbola with vertices and and a conjugate axis that measures 6 units.

    Part B: The Hyperbola in General Form

      Rewrite in standard form and graph.

      Given the general form, determine the intercepts.

      Find the equations of the asymptotes to the given hyperbola.

      Given the graph of a hyperbola, determine its equation in general form.

    Part C: Identifying the Conic Sections

      Identify the following as the equation of a line, parabola, circle, ellipse, or hyperbola.

      Identify the conic sections and rewrite in standard form.

    Part D: Discussion Board

    1. Develop a formula for the equations of the asymptotes of a hyperbola. Share it along with an example on the discussion board.

    2. Make up your own equation of a hyperbola, write it in general form, and graph it.

    3. Do all hyperbolas have intercepts? What are the possible numbers of intercepts for a hyperbola? Explain.

    4. Research and discuss real-world examples of hyperbolas.

Answers

  1. Center: ; ; ; opens left and right; vertices: ,

  2. Center: ; , ; opens upward and downward; vertices: ,

  3. Center: ; , ; opens upward and downward; vertices: ,

  4. x-intercepts: ; y-intercepts: none

  5. x-intercepts: none; y-intercepts:

  6. x-intercepts: ; y-intercepts: none

  7. x-intercepts: ; y-intercepts: none

  1. ;

  2. ;

  3. ;

  4. ;

  5. ;

  6. ;

  7. ;

  8. ;

  9. ;

  10. x-intercepts: , ; y-intercepts:

  11. x-intercepts: ; y-intercepts: none

  12. x-intercepts: , ; y-intercepts: ,

  1. Hyperbola;

  2. Ellipse;

    Sours: https://saylordotorg.github.io/text_intermediate-algebra/s11-04-hyperbolas.html

    Now discussing:

    And this has long been a painful problem for her. But back to the thoughts of our aunt. After a visit to her friend, she hurried to the market, where her new acquaintance, Pyotr Ivanovich, was selling in one of the shops. Seeing his sweetheart, Peter quickly served the last client and led her into the back room.

    Oh, Petenka, how I missed.



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