Chapter 9.6, Problem 10WE

To determine

To calculate: To find the equation of the hyperbola

Answer to Problem 10WE

Equation of hyperbola is $\frac{{\left(x-2\right)}^2}{9}-\frac{{\left(y-3\right)}^2}{40}=1$

Explanation of Solution

Given information: Foci is $\left(-5,3\right)$ and $\left(9,3\right)$ and difference of the focal radii is $6$

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:

  $\left(-5,3\right)$ and $\left(9,3\right)$

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

Thus,

  $\begin{array}{l}k=3\\ h-c=-5---(1)\\ h+c=9----(2)\end{array}$

On solving (1) and (2):

  $\begin{array}{l}2h=4\\ h=2\end{array}$

And

  $\begin{array}{l}2-c=-5\\ c=7\end{array}$

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

Thus, $2a=6$

On solving:

  $\begin{array}{l}a=\frac{6}{2}\\ a=3\end{array}$

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

  $\begin{array}{l}{b}^2=c^2-a^2\\ b^2=7^2-3^2\\ b^2=49-9\\ b^2=40\\ b=\pm 2\sqrt{10}\end{array}$

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

Substituting the values:

  $\begin{array}{l}\frac{{\left(x-2\right)}^2}{3^2}-\frac{{\left(y-3\right)}^2}{{\left(2\sqrt{10}\right)}^2}=1\\ \frac{{\left(x-2\right)}^2}{9}-\frac{{\left(y-3\right)}^2}{40}=1\end{array}$

Conclusion:

Hence, equation of hyperbola is $\frac{{\left(x-2\right)}^2}{9}-\frac{{\left(y-3\right)}^2}{40}=1$