Chapter 11.3, Problem 37E

To determine

To find: the equation for the hyperbola.

Answer to Problem 37E

The equation of hyperbola is $\frac{5y^2}{64}-\frac{5x^2}{256}=1$ .

Explanation of Solution

Given information:

From the figure, provided in the question it is observed that

foci of hyperbola : $\left(0,\pm 8\right)$

asymptotes of hyperbola : $y=\pm \frac{1}{2}x$

Calculation:

since from the given foci it is observed that the vertices of the hyperbola are on the y-axis, so the hyperbola has a vertical transverse axis.

So the equation is of the form,

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

For transverse axis hyperbola, the asymptote

  $y=\pm \frac{a}{b}x$

So from the $y=\pm \frac{1}{2}x$ asymptote it can be observed that ,

  $\begin{array}{l}\frac{a}{b}=\frac{1}{2}\\ 2a=b\end{array}$

Use the relation ,

  $\begin{array}{l}{c}^2=a^2+b^2\\ {\left(8\right)}^2={\left(a\right)}^2+{\left(2a\right)}^2\text{ [substitute c by 8 and b by 2a]}\\ {\text{64=a}}^2{\text{ +4a}}^2\text{ [square]}\\ 5a^2=64\text{ [add]}\\ a=\frac{8}{\sqrt{5}}\text{ [square root]}\end{array}$

Because to find value of $b$ substitute the $a=\frac{8}{\sqrt{5}}$ in $2a=b$ ,

  $\begin{array}{l}2a=b\\ b=2·\frac{8}{\sqrt{5}}\text{ [multiply]}\\ b=\frac{16}{\sqrt{5}}\end{array}$

so the equation of hyperbola is,

  $\begin{array}{l}\frac{y^2}{{\left(a\right)}^2}-\frac{x^2}{b^2}=1\\ \frac{y^2}{{\left(\frac{8}{\sqrt{5}}\right)}^2}-\frac{x^2}{{\left(\frac{16}{\sqrt{5}}\right)}^2}=1\text{ [substitute]}\\ \frac{5y^2}{64}-\frac{5x^2}{256}=1\text{ [simplify] }\end{array}$

Hence, the equation of hyperbola is $\frac{5y^2}{64}-\frac{5x^2}{256}=1$ .