Chapter 10.5, Problem 21PPSE

Solution

To find: The vertices, foci, and the asymptotes of the hyperbola $36x^2-8y^2=288$ . Also, sketch the graph.

The vertices of the hyperbola are $\left(\pm 2\sqrt{2},0\right)$ , foci are $\left(\pm 2\sqrt{11},0\right)$ and the asymptotes are $y=\pm \frac{3}{\sqrt{2}}x$

Given information: The vertices equation of the hyperbola is $36x^2-8y^2=288$

Formula used: The equation of the hyperbola,

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , here transverse axis is vertical.

Vertices of the hyperbola are $\left(\pm a,0\right)$ and the foci are $\left(\pm c,0\right)$ .

Here, $c^2=a^2+b^2$

Slope of asymptotes are $m=\pm \frac{b}{a}$ , then asymptotes are $y=\pm \frac{b}{a}x$

Calculation:

Consider the equation of the hyperbola as,

  $36x^2-8y^2=288$

This can be rewritten as,

  $\frac{36x^2}{288}-\frac{8y^2}{288}=1$

  $\frac{x^2}{8}-\frac{y^2}{36}=1$  ...... (1)

Compare equation (1) with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , then

  $a^2=8$ implies that $a=\pm 2\sqrt{2}$

And $b^2=36$ implies that $b=\pm 6$

Also, $c^2=a^2+b^2$ implies that,

  $\begin{array}{c}{c}^2=8+36\\ =44\\ c=\pm 2\sqrt{11}\end{array}$

Hence, the vertices of the hyperbola are $\left(\pm a,0\right)=\left(\pm 2\sqrt{2},0\right)$ , foci are $\left(\pm c,0\right)=\left(\pm 2\sqrt{11},0\right)$

Also, the asymptotes are

  $\begin{array}{c}y=\pm \frac{b}{a}x\\ =\pm \frac{6}{2\sqrt{2}}x\\ =\pm \frac{3}{\sqrt{2}}x\end{array}$

The graph of the hyperbola can be drawn as,