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Q. 59

Expert-verifiedFound in: Page 773

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.

We are given,

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

Now,

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

Hence proved.

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