If $e$ is eccentricity of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a b)$ and $e'$ is eccentricity of $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a

Question

If $e$ is eccentricity of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a > b)$ and $e'$ is eccentricity of $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a < b),$ then

Dr. Abubacker · Accepted Answer

(c)
$e^{2} = \frac{a^{2} - b^{2}}{a^{2}}$ and ${e'}^{2} = \frac{b^{2} - a^{2}}{b^{2}}$
$\Rightarrow \frac{1}{e^{2}} + \frac{1}{{e'}^{2}} = 1$

Q. If $e$ is eccentricity of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a > b)$ and $e'$ is eccentricity of $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a < b),$ then

Explanation

Reviewed by Dr. Abubacker

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