Graphing Hyperbolas

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 $\frac{x^2}{a^2} − \frac{y^2}{b^2}=1$ for horizontal hyperbolas and the standard form $\frac{y^2}{a^2} − \frac{x^2}{b^2}=1$ 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.
   1. if the equation is in the form $\frac{x^2}{a^2} − \frac{y^2}{b^2}=1$ then answer
      • the transverse axis is on the $x$-axis
      • the coordinates of the vertices are $±a,0)$
      • the coordinates of the co-vertices are $(0,±b)$
      • the coordinates of the foci are $(±c,0)$
      1. the equations of the asymptotes are $y=± \frac{b}{a}x$
   2. If the equation is in the form $\frac{y^2}{a^2} − \frac{x^2}{b^2}=1$, then
      • the transverse axis is on the $y$-axis
      • the coordinates of the vertices are $(0,±a)$
      • the coordinates of the co-vertices are $(±b,0)$
      • the coordinates of the foci are $(0,±c)$
      • the equations of the asymptotes are $y=± \frac{a}{b}x$
3. Solve for the coordinates of the foci using the equation $c=± \sqrt{a^2+b^2}$.
4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.

EXAMPLE 4

Graphing a Hyperbola Centered at $(0, 0)$ Given an Equation in Standard Form

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

Solution

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

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

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

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

$c=± \sqrt{a^2+b^2} = ± \sqrt{64+36} =± \sqrt{100}=±10$

Therefore, the coordinates of the foci are $(0,±10)$

The equations of the asymptotes are $y=±\frac{a}{b}x=±\frac{8}{6}x=±\frac{4}{3}x$

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 8.

Figure 8

TRY IT #4

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

Source: Rice University, https://openstax.org/books/college-algebra/pages/8-2-the-hyperbola
This work is licensed under a Creative Commons Attribution 4.0 License.

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 $\frac{(x−h)^2}{a^2}−\frac{(y−k)^2}{b2}=1$ for horizontal hyperbolas, and $\frac{(y−k)^2}{a^2} − \frac{(x−h)^2}{b^2}=1$ 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 $\frac{(x−h)^2}{a^2}− \frac{(y−k)^2}{b^2}=1$, then
      • the transverse axis is parallel to the $x$-axis
      • the center is $(h,k)$
      • the coordinates of the vertices are $(h±a,k)$
      • the coordinates of the co-vertices are $(h,k±b)$
      • the coordinates of the foci are $(h±c,k)$
      • the equations of the asymptotes are $y=±\frac{b}{a}(x−h)+k$
   2. If the equation is in the form $\frac{(y−k)^2}{a^2}− \frac{(x−h)^2}{b^2}=1$, then
      • the transverse axis is parallel to the $y$-axis
      • the center is $(h,k)$
      • the coordinates of the vertices are $(h,k±a)$
      • the coordinates of the co-vertices are $(h±b,k)$
      • the coordinates of the foci are $(h,k±c)$
      • the equations of the asymptotes are $y=±\frac{a}{b}(x−h)+k$
3. Solve for the coordinates of the foci using the equation $c=±\sqrt{a^2+b^2}$.
4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.

EXAMPLE 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.

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

The standard form that applies to the given equation is $\frac{(x−h)^2}{a^2} − \frac{(y−k)^2}{b^2}=1$, 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 $(h±a,k)=(2±6,−5)$, or $(−4,−5)$ and $(8,−5)$
• the coordinates of the co-vertices are $(h,k±b)=(2,−5±9)$, or $(2,−14)$ and $(2,4)$
• the coordinates of the foci are (h±c,k), where $c=± \sqrt{a^2+b^2}$. Solving for $c$, we have

$c = ± \sqrt{36+81} = ± \sqrt{117} = ± 3 \sqrt{13}$

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

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

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 9.

Figure 9

TRY IT #5

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

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. See Figure 10. For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide!

Figure 10 Cooling towers at the Drax power station in North Yorkshire, United Kingdom

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 6

Solving Applied Problems Involving Hyperbolas

The design layout of a cooling tower is shown in Figure 11. 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.

Figure 11 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.

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: $\frac{x^2}{a^2} − \frac{y^2}{b^2}=1$, 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 $60$ 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,

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

$\frac{x^2}{900} − \frac{y^2}{14400.3636} = 1$, or $\frac{x^2}{30^2} − \frac{y^2}{120.0015^2} =1$

TRY IT #6

A design for a cooling tower project is shown in Figure 12. 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.

Figure 12