Q. 59 - Calculus

Question

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

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

Step-by-Step Solution

Answer

Hence proved.

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,

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

Hence proved.