Chapter 8.3, Problem 41E

Solution

To calculate: To calculate the vertices and foci of the hyperbola.

The value of center is $\left(2,-3\right)$ .

The value of foci are $\left(2,-3+\sqrt{145}\right),\left(2,-3-\sqrt{145}\right)$ .

The value of vertices are $\left(0,5\right),\left(0,-5\right)$ .

Given information: The given equation of a hyperbola is:

  $\frac{{\left(y+3\right)}^2}{64}-\frac{{\left(x-2\right)}^2}{81}=1$ .

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.

  $\frac{{\left(y+3\right)}^2}{8^2}-\frac{{\left(x-2\right)}^2}{9^2}=1$

The vertices in given general form can be written as:

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

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

The values of $\left(a,b\right)=\left(9,8\right)$ .

Then, the vertices will be calculated as:

  $\begin{array}{c}Vertices=\left(2,-3+8\right),\left(2,-3-8\right)\\ =\left(2,5\right),\left(2,-11\right)\end{array}$

The foci can be calculated as:

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

The value of

  $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{l}c=\sqrt{64+81}\\ c=\sqrt{145}\end{array}$

Thus, the value of foci is:

  $Foci=\left(2,\sqrt{145}-3\right),\left(2,-\sqrt{145}-3\right)$

Thus, the values are:

Foci are $=\left(2,\sqrt{145}-3\right),\left(2,-\sqrt{145}-3\right)$ .

Vertices are $\left(0,5\right),\left(0,-5\right)$ .

Center is $\left(2,-3\right)$ .