Chapter 9.3, Problem 25E

To determine

To find: The center, vertices, foci and asymptotes of the hyperbola equation $\frac{y^2}{16}-\frac{x^2}{4}=1$ and sketch its graph using the asymptotes as an aid and use graphing utility to verify you graph.

Answer to Problem 25E

The center is $\left(0,0\right)$ , vertices is $\left(0,\pm 4\right)$ , foci is $\left(0,\pm 2\sqrt{5}\right)$ and asymptotes is $y=\pm 2x$ .

Explanation of Solution

Given information:

The equation of hyperbola is $\frac{y^2}{16}-\frac{x^2}{4}=1$ .

Calculation:

Rewrite the equation in standard form as follows.

  $\frac{{\left(y-0\right)}^2}{4^2}-\frac{{\left(x-0\right)}^2}{2^2}=1$

Compare the equation $\frac{y^2}{16}-\frac{x^2}{4}=1$ with standard equation $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ .

Therefore, the center of the hyperbola is $\left(h,k\right)=\left(0,0\right)$ and $a=4,b=2$ .

The vertices $\left(h+a,k\right)$ and $\left(h-a,k\right)$ are the two bending points of the hyperbola with the center $\left(h,k\right)$ and the semi axis $a$ and $b$ .

Find the vertices as follows.

  $\begin{array}{c}\left(h,k+a\right)=\left(0,0+4\right)\\ =\left(0,4\right)\\ \left(h,k-a\right)=\left(0,0-4\right)\\ =\left(0,-4\right)\end{array}$

For right-left facing hyperbola, the foci are defined as $\left(h,k+c\right)$ and $\left(h,k-c\right)$ , where $c=\sqrt{a^2+b^2}$ is the distance from the center $\left(h,k\right)$ to a focus.

Calculate the value of $c$ by substituting the value of $a$ and $b$ in the equation $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{c}c=\sqrt{4^2+2^2}\\ =\sqrt{16+4}\\ =\sqrt{20}\\ =2\sqrt{5}\end{array}$

Find the focus point as follows.

  $\begin{array}{c}\left(h,k+c\right)=\left(0,0+2\sqrt{5}\right)\\ =\left(0,2\sqrt{5}\right)\\ \left(h,k-c\right)=\left(0,0-2\sqrt{5}\right)\\ =\left(0,-2\sqrt{5}\right)\end{array}$

Substitute 4 for $a$ , $2$ for $b$ , 0 for $h$ and 0 for $k$ in the equation $y=\pm \frac{a}{b}\left(x-h\right)+k$ to find the asymptotes.

  $\begin{array}{c}y=\pm \frac{4}{2}\left(x-0\right)+0\\ =\pm 2x\end{array}$

Sketch the graph with the asymptotes as shown below.

  

Verify the graph by using graphing utility as shown below.

  

Therefore, the center is $\left(0,0\right)$ , vertices is $\left(0,\pm 4\right)$ , foci is $\left(0,\pm 2\sqrt{5}\right)$ and asymptotes is $y=\pm 2x$ .