Chapter 9.5, Problem 10PPS

To determine

To Find: The equation of hyperbola

Answer to Problem 10PPS

  $\frac{{\left(x-3\right)}^2}{25}-\frac{{\left(y+6\right)}^2}{11}=1$

Explanation of Solution

Given:

  

Formula Used:

  $\begin{array}{l}\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1\\ \frac{y^2}{a^2}-\frac{x^2}{b^2}=1\\ y=\frac{b}{a}xor=-\frac{b}{a}x\\ e=\frac{\sqrt{a^2+b^2}}{a}\\ l=\frac{2b^2}{a}\end{array}$

Calculation:

  $\begin{array}{l}a=5\\ c=6\\ a^2+b^2=c^2\\ 25+b^2=36\\ b=\sqrt{36-25}=\sqrt{11}\\ \text{Horizontal Center}\\ \frac{-2+8}{2},\frac{-6-6}{2}=\left(3,-6\right)=\left(h,k\right)\\ \text{Equation of hyperbola}\\ \frac{{\left(x-3\right)}^2}{5^2}-\frac{{\left(y-\left(-6\right)\right)}^2}{{\left(\sqrt{11}\right)}^2}=1\\ \Rightarrow \frac{{\left(x-3\right)}^2}{25}-\frac{{\left(y+6\right)}^2}{11}=1\end{array}$

Conclusion:

Thus equation of given hyperbola is $\frac{{\left(x-3\right)}^2}{25}-\frac{{\left(y+6\right)}^2}{11}=1$ .