Chapter 9.6, Problem 9WE

To determine

To calculate: To find the equation of the hyperbola

Answer to Problem 9WE

Equation of hyperbola is $\frac{{\left(y+5\right)}^2}{4}-\frac{{\left(x-3\right)}^2}{5}=1$

Explanation of Solution

Given information: Foci is $\left(3,-8\right)$ and $\left(3,-2\right)$ and difference of the focal radii is $4$

Formula Used:

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

  $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ , where $b^2=c^2-a^2$

Equation of the asymptotes are $y=\pm \frac{b}{a}\left(x-h\right)+k$

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

  $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ , where $b^2=c^2-a^2$

Equation of the asymptotes are $y=\pm \frac{a}{b}\left(x-h\right)+k$

Calculation:

Foci of hyperbola are given as follows:

  $\left(3,-8\right)$ and $\left(3,-2\right)$

Comparing the foci with $\left(h,k\pm c\right)$

Thus,

  $\begin{array}{l}h=3\\ k-c=-8---(1)\\ k+c=-2----(2)\end{array}$

On solving (1) and (2):

  $\begin{array}{l}2k=-10\\ k=-5\end{array}$

And

  $\begin{array}{l}-5-c=-8\\ c=-3\end{array}$

Also given that difference of the focal radii is $6$

Thus, $2a=4$

On solving:

  $\begin{array}{l}a=\frac{4}{2}\\ a=2\end{array}$

Now, $b^2$ is calculated as follows:

  $\begin{array}{l}{b}^2=c^2-a^2\\ b^2=-3^2-2^2\\ b^2=9-5\\ b^2=5\\ b=\pm \sqrt{5}\end{array}$

Equation of hyperbola $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ :

Substituting the values:

  $\begin{array}{l}\frac{{\left(y+5\right)}^2}{2^2}-\frac{{\left(x-3\right)}^2}{{\left(\sqrt{5}\right)}^2}=1\\ \frac{{\left(y+5\right)}^2}{4}-\frac{{\left(x-3\right)}^2}{5}=1\end{array}$

Conclusion:

Hence, equation of hyperbola is $\frac{{\left(y+5\right)}^2}{4}-\frac{{\left(x-3\right)}^2}{5}=1$