Chapter 8.3, Problem 6E

Solution

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

The value of foci are $\left(\sqrt{13},0\right),\left(-\sqrt{13},0\right)$ .

The value of vertices are $\left(2,0\right),\left(-2,0\right)$ .

Given information: The given equation of a hyperbola is:

  $9x^2-4y^2=36$ .

Calculation:

The given equation can be written in the standard form as:

  $\begin{array}{l}\frac{9x^2}{36}-\frac{4y^2}{36}=\frac{36}{36}\\ \frac{x^2}{4}-\frac{y^2}{9}=1\end{array}$

The general form of hyperbola is:

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

So, after comparing, we get.

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

The vertices in given general form can be written as:

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

Here, $\left(h,k\right)=\left(0,0\right)$ .

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

Then, the vertices will be calculated as:

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

The foci can be calculated as:

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

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:

  $\begin{array}{c}Foci=\left(0+\sqrt{13},0\right),\left(0-\sqrt{13},0\right)\\ =\left(\sqrt{13},0\right),\left(-\sqrt{13},0\right)\end{array}$

Thus, the values are:

Foci are $\left(\sqrt{13},0\right),\left(-\sqrt{13},0\right)$ .

Vertices are $\left(2,0\right),\left(-2,0\right)$ .