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Question: If normal at any point P to the ellipse \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\), a \> b me...

If normal at any point P to the ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b meet the axes at M and N so that PMPN=23\frac{PM}{PN} = \frac{2}{3}, then value of eccentricity is

A

12\frac{1}{\sqrt{2}}

B

23\sqrt{\frac{2}{3}}

C

13\frac{1}{\sqrt{3}}

D

None of these

Answer

13\frac{1}{\sqrt{3}}

Explanation

Solution

∵\because PMPN=23⇒b2a2=23⇒1−e2=23⇒e2=13\frac{PM}{PN} = \frac{2}{3} \Rightarrow \frac{b^{2}}{a^{2}} = \frac{2}{3} \Rightarrow 1 - e^{2} = \frac{2}{3} \Rightarrow e^{2} = \frac{1}{3}

∴ e = x24−y29\frac{x^{2}}{4} - \frac{y^{2}}{9}