Chapter 9, Problem 10CR

To determine

To find: The equation of the ellipse.

Answer to Problem 10CR

The equation of the ellipse is $\frac{x^2}{12}+\frac{y^2}{16}=1$

Explanation of Solution

Given:

Foci: $F1\left(0,-2\right)andF2\left(0,2\right)$

sum of the focal radii is $8$ .

Calculation:

An ellipse is the set of all points in the planer, such that the sum of the distances from $P$ to two fixed points is a given constant. Each of the fixed points is called a focus of the ellipse. If the foci are $F1$ and $F2$ , then for each point of the ellipse, $PF1$ and $PF2$ are the focal radii of $P$ .

Suppose a point $P(x,y)$ is on the ellipse if and only if the following is true:

  $PF1+PF2=8$

  $\begin{array}{l}\sqrt{x^2+{\left(y-(-2)\right)}^2}+\sqrt{x^2+{\left(y-(2)\right)}^2}=8\\ \sqrt{x^2+{\left(y+2\right)}^2}+\sqrt{x^2+{\left(y-(2)\right)}^2}=8\\ \sqrt{x^2+{\left(y+2\right)}^2}=8-\sqrt{x^2+{\left(y-(2)\right)}^2}\end{array}$

Now squaring both side of the above equation,

  $\begin{array}{l}{\left(\sqrt{x^2+{\left(y+2\right)}^2}\right)}^2={\left(8-\sqrt{x^2+{\left(y-2\right)}^2}\right)}^2\\ x^2+{\left(y+2\right)}^2=64-16\sqrt{x^2+{\left(y-2\right)}^2}+x^2+{\left(y-2\right)}^2\\ x^2+y^2+4y+4=64-16\sqrt{x^2+{\left(y-2\right)}^2}+x^2+y^2-4y+4\\ 4y=64-16\sqrt{x^2+{\left(y-2\right)}^2}-4y\\ 16\sqrt{x^2+{\left(y-2\right)}^2}=64-8y\\ 2\sqrt{x^2+{\left(y-2\right)}^2}=8-y\left[Dividingby8\right]\\ {\left(2\sqrt{x^2+{\left(y-2\right)}^2}\right)}^2={\left(8-y\right)}^2\left[Squaringbothside\right]\\ 4\left(x^2+{\left(y-2\right)}^2\right)={\left(8-y\right)}^2\\ 4\left(x^2+y^2-4y+4\right)=64-16y+y^2\\ 4x^2+4y^2-16y+16=64-16y+y^2\\ 4x^2+4y^2+16=64+y^2\\ 4x^2+3y^2=48\\ 4\frac{x^2}{48}+3\frac{y^2}{48}=1\\ \frac{x^2}{12}+\frac{y^2}{16}=1\end{array}$

Conclusion:

Therefore, the equation of the ellipse is $c$. $\frac{x^2}{12}+\frac{y^2}{16}=1$