Chapter EP, Problem 10.5.7EP

To determine

To find:The graph, vertices, foci and asymptotes of hyperbola.

Answer to Problem 10.5.7EP

The graph, vertices, foci and asymptotes of hyperbola is determined.

Explanation of Solution

Given:

$x^2-9y^2=36$

Calculation:

The general form of the hyperbola is given as,

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

Write the given equation in general form.

$\frac{x^2}{36}-\frac{y^2}{4}=1$

Compare both equations,

$\begin{array}{l}{a}^2=36\\ a=6\\ b^2=4\\ b=2\end{array}$

The center of the hyperbola is $\left(h,k\right)$ is $\left(0,0\right)$

The vertices of the hyperbola is $\left(6,0\right)$ and $\left(-6,0\right)$

The coordinates is calculated as,

$\begin{array}{c}c=\sqrt{a^2+b^2}\\ =\sqrt{36+4}\\ =\sqrt{40}\end{array}$

The coordinates of the foci of the hyperbola is $\left(\sqrt{40},0\right)\text{ and }\left(-\sqrt{40},0\right)$

The asymptotes is calculated as,

$\begin{array}{c}y=\pm \frac{b}{a}x\\ =\pm \frac{1}{3}x\end{array}$

The graph is plotted as,

Thus, the graph, vertices, foci and asymptotes of hyperbola is determined.