Chapter 8.3, Problem 47E

Solution

To calculate: To calculate the vertices and foci , eccentricity and graph of an equation of the hyperbola.

The value of center is $\left(3,1\right)$ .

The value of foci are $\left(3,1+\sqrt{13}\right),\left(3,1-\sqrt{13}\right)$ .

The value of vertices are $\left(3,4\right),\left(3,-2\right)$ .

The value of eccentricity is $e=\frac{\sqrt{13}}{3}$ .

The graph of an equation is:

  

Given information: The given equation of a hyperbola is:

  $4{\left(y-1\right)}^2-9{\left(x-3\right)}^2=36$ .

Calculation:

The standard form of the hyperbola is:

  $\begin{array}{l}\frac{4{\left(y-1\right)}^2}{36}-\frac{9{\left(x-3\right)}^2}{36}=1\\ \frac{{\left(y-1\right)}^2}{9}-\frac{{\left(x-3\right)}^2}{4}=1\end{array}$

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.

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

The vertices in given general form can be written as:

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

Here, $\left(h,k\right)=\left(3,1\right)$ .

The values of $\left(a,b\right)=\left(2,3\right)$ .

Then, the vertices will be calculated as:

  $\begin{array}{c}Vertices=\left(3,1-3\right),\left(3,1+3\right)\\ =\left(3,-2\right),\left(3,4\right)\end{array}$

The foci can be calculated as:

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

The value of

  $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{l}c=\sqrt{9+4}\\ c=\sqrt{13}\end{array}$

Thus, the value of foci is:

  $Foci=\left(3,1+\sqrt{13}\right),\left(3,1-\sqrt{13}\right)$

The eccentricity of a hyperbola is:

  $\begin{array}{l}e=\frac{c}{b}\\ e=\frac{\sqrt{13}}{3}\end{array}$

The graph of an equation is:

  

Thus, the values are:

Foci are $=\left(3,1+\sqrt{13}\right),\left(3,1-\sqrt{13}\right)$ .

Vertices are $\left(3,-2\right),\left(3,4\right)$ .

Center is $\left(3,1\right)$ .

Eccentricity is $e=\frac{\sqrt{13}}{3}$ .