If normal at any point $P$ to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a b)$

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If normal at any point $P$ to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a > b)$ meet the axes at $M$ and $N$ so that $\frac{PM}{PN} = \frac{2}{3}$, then the value of eccentricity is:

Accepted Answer

Sol.(c) $(PM)^{2} = a^{2}\cos^{2}	heta\left( 1 - e^{2} ight)^{2} + b^{2}\sin^{2}	heta\ = b^{2}\left( 1 - e^{2}\cos^{2}	heta ight)$$$\begin{matrix} & \ (PN)^{2} = & a^{2}\cos^{2}	heta + \frac{\sin^{2}	heta}{b^{2}}\left( b^{2} + a^{2}e^{2} ight)^{2} \ = & \frac{a^{4}}{b^{2}}\left( 1 - e^{2}\cos^{2}	heta ight) \end{matrix}$$

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