Chapter 9.5, Problem 3CYU

To determine

To Find: The equation of given hyperbola.

Answer to Problem 3CYU

  $\frac{x^2}{64}-\frac{y^2}{25}=1$

Explanation of Solution

Given:

  

Formula Used:

  $\begin{array}{l}{a}^2+b^2=c^2\\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1Horizontal\\ \frac{y^2}{a^2}-\frac{x^2}{b^2}=1Vertical\end{array}$

Calculation:

  $\begin{array}{l}\left(h,k\right)=\left(0,0\right)\\ b=5\\ a=8\\ \\ \frac{x^2}{64}-\frac{y^2}{25}=1\end{array}$

Conclusion:

Thus , the equation of given hyperbola is $\frac{x^2}{64}-\frac{y^2}{25}=1$