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Q. 59

Expert-verified
Found in: Page 773

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

Prove that for an ellipse or a hyperbola the eccentricity is given by $e=\frac{\text{the distance between the foci}}{\text{the distance between the vertices}}$

Hence proved.

See the 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,

$\text{Let the foci of the ellipse are}\left(±\sqrt{{A}^{2}-{B}^{2}},0\right)\text{.}\phantom{\rule{0ex}{0ex}}\text{Formula for the distance}=\sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\text{Distance}=\sqrt{{\left(\sqrt{{A}^{2}-{B}^{2}}-\left(-\sqrt{{A}^{2}-{B}^{2}}\right)\right)}^{2}+\left(0-0{\right)}^{2}}\phantom{\rule{0ex}{0ex}}\text{Distance}=\sqrt{{\left(2\sqrt{{A}^{2}-{B}^{2}}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\text{Thus the distance between the foci is,}\phantom{\rule{0ex}{0ex}}\text{Distance}=2\sqrt{{A}^{2}-{B}^{2}}\phantom{\rule{0ex}{0ex}}Wehavetheeccentricitye=\frac{\sqrt{{A}^{2}-{B}^{2}}}{A}\phantom{\rule{0ex}{0ex}}\text{Then}e=\frac{2\sqrt{{A}^{2}-{B}^{2}}}{2A}\phantom{\rule{0ex}{0ex}}\text{Thus,}e=\frac{\text{the distance between foci}}{\text{the distance between the vertices}}$

Hence proved.