Chapter 9.4, Problem 35WE

To determine

To find: an equation of the ellipse.

Answer to Problem 35WE

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

Explanation of Solution

Given:

The foci: $F_1\left(-c,0\right)$ and $F_2\left(c,0\right)$

The sum of the focal radii is $2a$ .

Calculation:

An ellipse is the set f all points P in a plane such that the sum of the distances from P to two fixed points $F_1$ and $F_2$ is a given constant.

In order to find an equation of the given ellipse, use the distance formula.

By the definition, a point $P\left(x,y\right)$ is on the ellipse if and only if $P{F}_1+P{F}_2=2a$ .

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

Square both the sides and simplify.

  $\begin{array}{l}{\left(\sqrt{{\left(x+c\right)}^2+y^2}\right)}^2={\left(2a-\sqrt{{\left(x-c\right)}^2+y^2}\right)}^2\\ {\left(x+c\right)}^2+y^2={\left(2a\right)}^2-2\cdot 2a\sqrt{{\left(x-c\right)}^2+y^2}+{\left(\sqrt{{\left(x-c\right)}^2+y^2}\right)}^2\\ x^2+2xc+e^2+y^2=4a^2-4a\sqrt{{\left(x-c\right)}^2+y^2}+x^2-2xc+e^2+y^2\end{array}$

Combine the similar terms and rearrange.

  $\begin{array}{l}4xc=4a^2-4a\sqrt{\left(x-c\right)+y^2}\\ 4a\sqrt{{\left(x–c\right)}^2+y^2}=4a^2-4xc\end{array}$

Square both the sides again and simplify.

  $\begin{array}{l}{\left(4a\sqrt{{\left(x–c\right)}^2+y^2}\right)}^2={\left(4a^2-4xc\right)}^2\hfill \\ \begin{array}{l}16a^2\left({\left(x–c\right)}^2+y^2\right)={\left(4a\right)}^2–2\cdot 4a^2\cdot 4xc+{\left(4xc\right)}^2\\ 16a^2\left(x^2-2xc+c^2+y^2\right)=16a^2-32a^{2}xc+16x^2{c}^2\end{array}\hfill \end{array}$

Divide both the sides by 16.

  $\begin{array}{l}{a}^2\left(x^2-2xc+c^2+y^2\right)=a^2-2a^{2}xc+x^2{c}^2\\ a^2{x}^2-2a^{2}xc+a^2{c}^2+a^2{y}^2=a^2-2a^{2}xc+x^2{c}^2\end{array}$

Combine the similar terms and rearrange.

  $\begin{array}{l}{a}^2{x}^2-x^2{c}^2+a^2{y}^2=a^4-a^2{c}^2\\ x^2\left(a^2-c^2\right)+a^2{y}^2=a^2\left(a^2-c^2\right)\end{array}$

Substitute $b^2$ for $a^2-c^2$ .

  $x^2{b}^2+a^2{y}^2=a^2{b}^2$

Divide both the sides by $a^2{b}^2$ .

  $\begin{array}{l}\frac{x^2{b}^2}{a^2{b}^2}+\frac{a^2{y}^2}{a^2{b}^2}=1\\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\end{array}$

Conclusion:

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