Chapter 14.2, Problem 71SR

To determine

To Prove :Vertices, foci, equation of asymptotes and graph of hyperbola.

Answer to Problem 71SR

  $\begin{array}{l}Vertices(1,-6+2\sqrt{5}),(1,-6-2\sqrt{5})\\ Foci(1,-6+3\sqrt{5}),(1,-6-3\sqrt{5})\\ \text{Equation of asymptote, }y=\pm \frac{2}{\sqrt{5}}(x-1)-6\end{array}$

Explanation of Solution

Given information :

We know that, $\frac{{(y+6)}^2}{20}-\frac{{(x-1)}^2}{25}=1$

Calculation :

  $\begin{array}{l}Atx=1\\ \frac{{(y+6)}^2}{20}-\frac{{(1-1)}^2}{25}=1\\ $ {(y+6)}^2=20$y+6=\pm 2\sqrt{5}\\ y=-6\pm 2\sqrt{5}\\ \text{Vertices are, }(1,-6+2\sqrt{5}),(1,-6-2\sqrt{5})\\ For\text{ foci, }c=\pm \sqrt{a^2+b^2}=\pm \sqrt{20+25}=\pm \sqrt{45}=\pm 3\sqrt{5}\\ \text{Coordinates of foci, }(1,-6+3\sqrt{5}),(1,-6-3\sqrt{5})\\ Equationofasymptote,\\ y=\pm \frac{2\sqrt{5}}{5}(x-1)+(-6)=\pm \frac{2}{\sqrt{5}}(x-1)-6\end{array}$