Chapter 11.4, Problem 14E

To determine

To find: the center, foci, vertices and asymptotes of the hyperbola then sketch the graph

Answer to Problem 14E

The Center is $\left(8,-6\right)$ , vertices is $\left(9,-6\right),\left(7,-6\right)$ , foci is $\left(\sqrt{2}+8,-6\right)\left(-\sqrt{2}+8,-6\right)$ and the asymptote is $y=-6\left(x-8\right)$

Explanation of Solution

Given information: The question is

  ${\left(x-8\right)}^2-{\left(y+6\right)}^2=1$

Calculation: for the center, foci, vertices and asymptotes

  $\begin{array}{l}\frac{{\left(x+1\right)}^2}{9}-\frac{{\left(y-3\right)}^2}{16}=1\\ \Rightarrow \frac{{\left(x+1\right)}^2}{3^2}-\frac{{\left(y-3\right)}^2}{4^2}=1\\ \text{from standard equation:}\\ \frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1\\ a=3;b=4;c=5;h=-1;k=3\end{array}$

Center $\left(h,k\right)=\left(8,-6\right)$

Vertices $\left(h+a,k\right)\&\left(h-a,k\right)=\left(9,-6\right),\left(7,-6\right)$

Foci $\left(h+c,k\right)\left(h-c,k\right)=\left(\sqrt{2}+8,-6\right)\left(-\sqrt{2}+8,-6\right)$

Asymptote $y=k\frac{b}{a}\left(x-h\right)=y=-6\left(x-8\right)$