MA001: College Algebra

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.