Chapter 9.5, Problem 15WE

To determine

To calculate: To find the equation of hyperbola

Answer to Problem 15WE

Equation of hyperbola is $\frac{y^2}{9}-\frac{x^2}{4}=1$

Explanation of Solution

Given information: Asymptotes $y=\frac{3}{2}x$ and $y=-\frac{3}{2}x$ and Foci are $\left(0,-\text{}\sqrt{13}\right)$ and $\left(0,\text{}\sqrt{13}\right)$

Formula Used:

The hyperbola with center $\left(0,0\right)$ and foci $\left(-c,0\right)$ and $\left(c,0\right)$ and difference of focal radii $2a$ has the equation:

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

Equation of the asymptotes are $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$

The hyperbola with center $\left(0,0\right)$ and foci $\left(0,-c\right)$ and $\left(0,c\right)$ and difference of focal radii $2a$ has the equation:

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

Equation of the asymptotes are $y=\frac{a}{b}x$ and $y=-\frac{a}{b}x$

Calculation:

Given that Foci of hyperbola are

  $\left(0,-\text{}\sqrt{13}\right)$ and $\left(0,\text{}\sqrt{13}\right)$

Thus,

  $c=\pm \sqrt{13}$

Also, Asymptotes are given as follows:

  $y=\frac{3}{2}x$ and $y=-\frac{3}{2}x$

Thus,

  $\begin{array}{l}a=3\\ b=2\end{array}$

Standard form of the equation of hyperbola is given as follows:

  $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

Substituting the value:

  $\begin{array}{l}\frac{y^2}{3^2}-\frac{x^2}{2^2}=1\\ \frac{y^2}{9}-\frac{x^2}{4}=1\end{array}$

Conclusion:

Hence, equation of hyperbola is $\frac{y^2}{9}-\frac{x^2}{4}=1$