Chapter 9.2, Problem 71E

To determine

To prove: The equation $a^2=b^2+c^2$ for the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a>0,b>0$ and the distance from the center of the ellipse $\left(0,0\right)$ to a focus is $c$ .

Explanation of Solution

Given information:

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a>0,b>0$ and the distance from the center of the ellipse $\left(0,0\right)$ to a focus is $c$ .

Calculation:

Plot $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ using graphing utility. Let $P$ is any point on the ellipse and $F_1{\text{ and F}}_2$ are the foci of the ellipse.

  

By definition, an ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant so that $P{F}_1+P{F}_2=\text{constant}$ .

From figure 1, use Pythagoras theorem to calculate $P{F}_1{\text{ and PF}}_2$ .

  ${\text{PF}}_{\text{2}}=\sqrt{b^2+c^2}$.........(1)

  $\begin{array}{l}{\text{PF}}_{\text{1}}=\sqrt{b^2+{\left(-c\right)}^2}\\ =\sqrt{b^2+c^2}\end{array}$.........(2)

Add equation (1) and (2):

  $\begin{array}{l}{\text{PF}}_{\text{1}}+{\text{PF}}_{\text{2}}=\sqrt{b^2+c^2}+\sqrt{b^2+c^2}\\ =2\sqrt{b^2+c^2}\end{array}$

  ${\text{PF}}_{\text{1}}+{\text{PF}}_{\text{2}}=2\sqrt{b^2+c^2}$.........(3)

Let us suppose $P$ moves from $(0,b)$ to $(a,0)$ as it can be any point on the ellipse.

  

From figure 2, use Pythagoras theorem to calculate $P{F}_1{\text{ and PF}}_2$ .

  ${\text{PF}}_{\text{2}}=a-c$.........(4)

  ${\text{PF}}_{\text{1}}=a+c$.........(5)

Add equation (4) and (5):

  $\begin{array}{l}{\text{PF}}_{\text{1}}+{\text{PF}}_{\text{2}}=a+\cancel{c}+a-\cancel{c}\\ =2a\end{array}$

  ${\text{PF}}_{\text{1}}+{\text{PF}}_{\text{2}}=2a$.........(6)

By equation (3) and (6):

  $\begin{array}{l}\cancel{2}a=\cancel{2}\sqrt{b^2+c^2}\\ a=\sqrt{b^2+c^2}\\ a^2=b^2+c^2\end{array}$

Therefore, $a^2=b^2+c^2$ for the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a>0,b>0$ and the distance from the center of the ellipse $\left(0,0\right)$ to a focus is $c$ has been proved.