Chapter 9, Problem 6CT

To determine

To find: The equation of the ellipse.

Answer to Problem 6CT

The equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$

Explanation of Solution

Given:

Foci : $F1\left(\sqrt{5},0\right)andF2\left(-\sqrt{5},0\right)$

sum of the focal radii is $6$

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:

  $PF1+PF2=6$

  $\begin{array}{l}\sqrt{{\left(x-\sqrt{5}\right)}^2+{\left(y\right)}^2}+\sqrt{{\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2}=6\\ \sqrt{{\left(x-\sqrt{5}\right)}^2+{\left(y\right)}^2}=6-\sqrt{{\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2}\end{array}$

Now squaring both side of the above equation,

  $\begin{array}{l}{\left[\sqrt{{\left(x-\sqrt{5}\right)}^2+{\left(y\right)}^2}\right]}^2={\left[6-\sqrt{{\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2}\right]}^2\\ {\left(x-\sqrt{5}\right)}^2+{\left(y\right)}^2=36-12\sqrt{{\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2}+{\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2\\ x^2-2\sqrt{5}x+5+y^2=36-12\sqrt{{\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2}+x^2+2\sqrt{5}x+5+y^2\\ -2\sqrt{5}x=36-12\sqrt{{\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2}+2\sqrt{5}x\\ 12\sqrt{{\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2}=36+4\sqrt{5}x\\ 3\sqrt{{\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2}=9+\sqrt{5}x\left[Dividingbothsideby4\right]\end{array}$

Again squaring both side of the above equation,

  $\begin{array}{l}{\left[3\sqrt{{\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2}\right]}^2={\left(9+\sqrt{5}x\right)}^2\\ 9\left({\left(x+\sqrt{5}\right)}^2+{\left(y\right)}^2\right)={\left(9+\sqrt{5}x\right)}^2\\ 9\left(x^2+2\sqrt{5}x+5+y^2\right)=\left(81+18\sqrt{5}x+5x^2\right)\\ 9x^2+18\sqrt{5}x+45+9y^2=81+18\sqrt{5}x+5x^2\\ 9x^2+45+9y^2=81+5x^2\\ 4x^2+9y^2=36\\ \frac{4x^2}{36}+\frac{4x^2}{36}=1\\ \frac{x^2}{9}+\frac{y^2}{4}=1\end{array}$

Conclusion:

Therefore, the equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$