Chapter 9.4, Problem 34WE

To determine

To prove: $b^2=a^2-c^2$ for the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci at $\left(-c,0\right)$ and $\left(c,0\right)$ .

Explanation of Solution

Proof:

The given equation of the ellipse is

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Show that $b^2=a^2-c^2$ if the foci are located at $\left(-c,0\right)$ and $\left(c,0\right)$ .

Take an arbitrary point P in a graph to prove the equation. The graph is as follows:

  

Suppose that, $F_1$ and $F_2$ are the foci and P is any point on the ellipse. Thus, $F_1=\left(c,0\right)$ and $F_2=\left(-c,0\right)$ .

According to the definition of the ellipse, if the foci are $F_1$ and $F_2$ , then for each point P of the ellipse, $P{F}_1$ and $P{F}_2$ are the focal radii of P and sum of $P{F}_1$ and $P{F}_2$ is a given constant.

Now take the point $\left(-a,0\right)$ as P and a constant k .

Then according to the definition,

  $P{F}_1+P{F}_2=k\mathrm{......}\left(1\right)$

By using distance formula,

  $\begin{array}{l}P{F}_1=\sqrt{{\left(-a-c\right)}^2+{\left(0-0\right)}^2}\hfill \\ =a+c\hfill \end{array}$

And,

  $\begin{array}{l}P{F}_2=\sqrt{{\left(-a+c\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{{\left(-\left(a-c\right)\right)}^2}\\ =a-c\end{array}$

By putting these in equation (1),

  $\begin{array}{l}P{F}_1+P{F}_2=k\hfill \\ a+c+a-c=k\hfill \\ k=2a\hfill \end{array}$

Therefore, $P{F}_1+P{F}_2=2a$

Again take the point $\left(0,b\right)$ as the point P .

Then according to the definition of the ellipse,

  $P{F}_1+P{F}_2=2a\left[\text{Since }k=2a\right]\mathrm{......}\left(2\right)$

By using distance formula,

  $P{F}_1=\sqrt{c^2+b^2}$

And,

  $P{F}_2=\sqrt{c^2+b^2}$

From equation (2),

  $\begin{array}{l}P{F}_1+P{F}_2=2a\\ 2\sqrt{c^2+b^2}=2a\end{array}$

By cancelling 2 from both sides,

  $\sqrt{c^2+b^2}=a$

By squaring both sides,

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

Conclusion:

Therefore, $b^2=a^2-c^2$ proved.