Eccentricity of ellipse whose latus rectum is equal to distance between two focus points, is

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10-2. Parabola, Ellipse, Hyperbola

easy

Eccentricity of the ellipse whose latus rectum is equal to the distance between two focus points, is

A

$\frac{{\sqrt 5 + 1}}{2}$

B

$\frac{{\sqrt 5 - 1}}{2}$

C

$\frac{{\sqrt 5 }}{2}$

D

$\frac{{\sqrt 3 }}{2}$

Solution

(b)$\frac{{2{b^2}}}{a} = 2ae$

${b^2} = {a^2}e$ or $e = \frac{{{b^2}}}{{{a^2}}}$

Also $e = \sqrt {1 – \frac{{{b^2}}}{{{a^2}}}} $or ${e^2} = 1 – e$ or ${e^2} + e – 1 = 0$

Therefore $e = \frac{{ – 1 \pm \sqrt 5 }}{2}$. As $e < 1,$ $e = \frac{{\sqrt 5 – 1}}{2}$.

Standard 11

Mathematics

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