Chapter 9, Problem 9CT

To determine

To find: the equation of ellipse using given foci and focal radii.

Answer to Problem 9CT

The equation for ellipse will be $\frac{{\left(x-2\right)}^2}{16}+\frac{y^2}{12}=1$ .

Explanation of Solution

Given:

The equation of an ellipse having foci $\left(0,0\right)$ and $\left(4,0\right)$ .

The sum of focal radii $8$ .

Concept used:

The equation of an ellipse:

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

Relation between foci and axis.

  $b^2=a^2-c^2.$

Calculation:

According to the given:

The equation of an ellipse having foci $\left(0,0\right)$ and $\left(4,0\right)$ .

The sum of focal radii $8$ .

The equation of an ellipse:

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

The center is the middle of the foci, which is $\left(2,0\right)$ .

  $\left(\frac{0+4}{2},0\right)=\left(2,0\right)$ .

  $c=2\text{ and }\left(h,k\right)=\left(2,0\right)$ .

  $a$ is the distance from the center to one of the vertices on the left or right side so, the sum of focal radii is $2a$ .

  $\begin{array}{l}{b}^2=a^2-c^2.\\ \Rightarrow 4^2-2^2=12.\\ b^2=12.\end{array}$

By putting the value sin equation:

  $\begin{array}{l}\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y-0\right)}^2}{12}=1.\\ \frac{{\left(x-2\right)}^2}{16}+\frac{y^2}{12}=1.\end{array}$

Hence, the equation for ellipse will be $\frac{{\left(x-2\right)}^2}{16}+\frac{y^2}{12}=1$ .