Chapter 8.3, Problem 44E

Solution

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

The value of center is $\left(0,0\right)$ .

The value of foci are $=\left(5\sqrt{2},0\right),\left(-5\sqrt{2},0\right)$ .

The value of vertices are $=\left(\sqrt{30},0\right),\left(-\sqrt{30},0\right)$ .

The graph of an equation is:

  

Given information: The given equation of a hyperbola is:

  $\frac{{\left(x\right)}^2}{30}-\frac{{\left(y\right)}^2}{20}=1$ .

Calculation:

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.

  $\frac{{\left(x\right)}^2}{30}-\frac{{\left(y\right)}^2}{20}=1$

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 $\left(a,b\right)=\left(\sqrt{30},\sqrt{20}\right)$ .

Then, the vertices will be calculated as:

  $\begin{array}{c}Vertices=\left(0+\sqrt{30},0\right),\left(0-\sqrt{30},0\right)\\ =\left(\sqrt{30},0\right),\left(-\sqrt{30},0\right)\end{array}$

The foci can be calculated as:

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

The value of

  $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{l}c=\sqrt{30+20}\\ c=\sqrt{50}\\ c=5\sqrt{2}\end{array}$

Thus, the value of foci is:

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

Thus, the values are:

Foci are $=\left(5\sqrt{2},0\right),\left(-5\sqrt{2},0\right)$ .

Vertices are $=\left(\sqrt{30},0\right),\left(-\sqrt{30},0\right)$ .

Center is $\left(0,0\right)$ .

The graph of an equation is: