Question

Prove that for an ellipse or a hyperbola the eccentricity is given by

$e=\frac{\text{the distance between the foci}}{\text{the distance between the vertices}}$

Short answer

Hence proved.

Step-by-step solution

Step 1. Given information.

We are given,

$e=\frac{\text{the distance between the foci}}{\text{the distance between the vertices}}$

Step 2. Explanation.

Now,

$$\begin{gathered} \text{Let the foci of the ellipse are}\left(\pm \sqrt{A^2-B^2},0\right)\text{.} \\ \text{Formula for the distance}=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2} \\ \text{Distance}=\sqrt{{\left(\sqrt{A^2-B^2}-\left(-\sqrt{A^2-B^2}\right)\right)}^2+(0-0{)}^2} \\ \text{Distance}=\sqrt{{\left(2\sqrt{A^2-B^2}\right)}^2} \\ \text{Thus the distance between the foci is,} \\ \text{Distance}=2\sqrt{A^2-B^2} \\ Wehavetheeccentricitye=\frac{\sqrt{A^2-B^2}}{A} \\ \text{Then}e=\frac{2\sqrt{A^2-B^2}}{2A} \\ \text{Thus,}e=\frac{\text{the distance between foci}}{\text{the distance between the vertices}} \end{gathered}$$

Hence proved.