Chapter 9.3, Problem 42E

To determine

To find:The standard form of the equation of hyperbola when the vertices are $\left(0,\pm 3\right)$ and asymptotes is $y=\pm 3x$ .

Answer to Problem 42E

The standard form of the equation of hyperbola is $\frac{y^2}{3^2}-\frac{x^2}{1^2}=1$ .

Explanation of Solution

Given information:

The vertices of hyperbola are $\left(0,\pm 3\right)$ and asymptotes of hyperbola is $y=\pm 3x$

Calculation:

The vertices of the hyperbola is given by $\left(h,k+a\right)$ and $\left(h,k-a\right)$ .Since the given vertices are $\left(0,-3\right)$ and $\left(0,3\right)$ . So by comparing it we get,

  $a=3$ , $h=0$ and $k=0$

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 $y=\pm 3x$ .So by comparing it with the above equation, we get

  $a=3$ , $b=1$ , $h=0$ and $k=0$ .

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 $a=3$ , $b=1$ , $h=0$ and $k=0$ to find the standard equation of the hyperbola

  $\begin{array}{c}\frac{{\left(y-0\right)}^2}{3^2}-\frac{{\left(x-0\right)}^2}{1^2}=1\\ \frac{y^2}{9}-\frac{x^2}{1}=1\end{array}$

Therefore, the standard form of the equation of hyperbola is $\frac{y^2}{3^2}-\frac{x^2}{1^2}=1$