Chapter 8.3, Problem 9E

Solution

To calculate: To draw the graph of the hyperbola.

The graph of the given equation is :

  

Given information: The given equation of a hyperbola is:

  $\frac{{\left(y-2\right)}^2}{4}-\frac{{\left(x+3\right)}^2}{16}=1$ .

Calculation:

The general form of hyperbola is:

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

So, after comparing, we get.

  $\frac{{\left(y-2\right)}^2}{{\left(2\right)}^2}-\frac{{\left(x+3\right)}^2}{{\left(4\right)}^2}=1$

The center and vertices in given general form can be written as:

Center $\left(h,k\right)=\left(-3,2\right)$

Vertices $=\left(h,k-b\right),\left(h,k+b\right)$ .

The values of $\left(a,b\right)=\left(4,2\right)$ .

Then, the vertices will be calculated as:

  $\begin{array}{c}Vertices=\left(-3,2+2\right),\left(-3,2-2\right)\\ =\left(-3,4\right),\left(3,0\right)\end{array}$

The foci can be calculated as:

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

The value of

  $c=\sqrt{a^2+b^2}$ .

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

Thus, the value of foci is:

  $Foci=\left(-3,2\sqrt{5}+2\right),\left(-3,-2\sqrt{5}+2\right)$

Thus, the values are:

Foci are $=\left(-3,2\sqrt{5}+2\right),\left(-3,-2\sqrt{5}+2\right)$ .

Vertices are $\left(-3,4\right),\left(3,0\right)$ .

The graph of the given equation is:

  

Thus, The correct option is A.