Question

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

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{\frac{2}{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{2}{3}$

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}$

Equation of any normal to the ellipse on a point $P(a\cos \theta ,b\sin \theta )$ is:

$\frac{ax}{cos\theta }-\frac{by}{sin\theta }=ae$

Then, $M(e\cos \theta ,0)$ and $N(0,\frac{ae\sin \theta }{b})$

Given: $\frac{PM}{PN}=\frac{2}{3}$

$\Rightarrow \frac{PN+MN}{PN}=\frac{2}{3}$

$\Rightarrow \frac{MN}{PN}=\frac{-1}{3}$

Point N divides the line PM in the ratio $1:3$

The coordinates of N can be equated to:

$0=\frac{-a\cos \theta +3e\cos \theta }{2}$

$\Rightarrow a=3e$ ----------(1)

And, $\frac{ae\sin \theta }{b}=\frac{|-b\sin \theta |}{2}$

$\Rightarrow b^2=2ae$

Using equation 1 we get:

$b^2=6e^2$ and $b^2=a^2-{\left(ae\right)}^2$

Using the above equations we get:

$6e^2=9e^2-9e^4$

$\Rightarrow 9e^4=3e^2$

$\Rightarrow e=\sqrt{\frac{1}{3}}$