Chapter 9.3, Problem 43E

To determine

To find:The standard form of the equation of hyperbola when the foci are $\left(0,\pm 8\right)$ and asymptotes is $y=\pm 4x$ .

Answer to Problem 43E

The standard form of the equation of hyperbola is $\frac{17y^2}{1024}-\frac{17x^2}{64}=1$ .

Explanation of Solution

Given information:

Thefoci of hyperbola are $\left(0,\pm 8\right)$ and asymptotes of hyperbola is $y=\pm 4x$ .

Calculation:

The asymptotes of the hyperbola is given by $y=\pm \frac{a}{b}\left(x-h\right)+k$ . The asymptotes of the hyperbola given is $y=\pm 4x$ .So by comparing it with the above equation, we get

  $a=4n$ , $b=1n$ , $h=0$ and $k=0$ .

The foci of the hyperbola is given by $\left(h,k+c\right)$ and $\left(h,k-c\right)$ , where $c^2=a^2+b^2$ . The foci of the hyperbola given is $\left(0,-8\right)$ and $\left(0,8\right)$ .

Calculate the value of $c$ by putting $k=0$ .

  $c=8$

Use $c^2=a^2+b^2$ to determine the value of $n$ .

  $\begin{array}{c}{c}^2=a^2+b^2\\ 8^2={\left(4n\right)}^2+{\left(n\right)}^2\\ 64=16n^2+n^2\\ 64=17n^2\end{array}$

Taking square root both the sides.

  $\begin{array}{c}\sqrt{17}n=\pm 8\\ n=\pm \frac{8}{\sqrt{17}}\end{array}$

Substitute $n=\frac{8}{\sqrt{17}}$ to find the value of $a$ and $b$ .

  $\begin{array}{l}a=4n\\ a=4\left(\frac{8}{\sqrt{17}}\right)\\ a=\frac{32}{\sqrt{17}}\end{array}$

and,

  $\begin{array}{l}b=n\\ b=\frac{8}{\sqrt{17}}\end{array}$

The standard equation of the hyperbola is given as $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$

Substitute $a=\frac{32}{\sqrt{17}}$ , $b=\frac{8}{\sqrt{17}}$ , $h=0$ and $k=0$ in the standard equation of the hyperbola.

  $\begin{array}{c}\frac{{\left(y-0\right)}^2}{{\left(\frac{32}{\sqrt{17}}\right)}^2}-\frac{{\left(x-0\right)}^2}{{\left(\frac{8}{\sqrt{17}}\right)}^2}=1\\ \frac{y^2}{\left(\frac{1024}{17}\right)}-\frac{x^2}{\left(\frac{64}{17}\right)}=1\\ \frac{17y^2}{1024}-\frac{17x^2}{64}=1\end{array}$

Therefore, the standard form of the equation of hyperbola is $\frac{17y^2}{1024}-\frac{17x^2}{64}=1$ .