Chapter 11.4, Problem 15E

To determine

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

Answer to Problem 15E

The Center is $\left(-1,0\right)$ , vertices is $\left(-1,\sqrt{5}\right),\left(-1,-\sqrt{5}\right)$ , foci is $\left(-1,1\right)\left(-1,-1\right)$ and the asymptote is $y=\pm \frac{1}{2}\left(x+1\right)$

Explanation of Solution

Given information: The question is

  $y^2-\frac{{\left(x+1\right)}^2}{4}=1$

Calculation: for the center, foci, vertices and asymptotes

  $\begin{array}{l}{y}^2-\frac{{\left(x+1\right)}^2}{4}=1\\ \Rightarrow y^2-\frac{{\left(x+1\right)}^2}{4}=1\\ \text{from standard equation:}\\ \frac{{\left(y\right)}^2}{1}-\frac{{\left(x+1\right)}^2}{2^2}=1\\ a=1;b=2;c=\sqrt{5};h=-1;k=0\end{array}$

Center $\left(h,k\right)=\left(-1,0\right)$

Vertices $\left(h+a,k\right)\&\left(h-a,k\right)=\left(-1,\sqrt{5}\right),\left(-1,-\sqrt{5}\right)$

Foci $\left(h+c,k\right)\left(h-c,k\right)=\left(-1,1\right)\left(-1,-1\right)$

Asymptote $y=k\frac{b}{a}\left(x-h\right)=y=\pm \frac{1}{2}\left(x+1\right)$