Question

The normal at any point $P$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ meets the x-axis at $A$ and y-axis at $B$. If $PA:PB=1:4,$ then the eccentricity of the ellipse is

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{4}$
• D. $\frac{\sqrt{3}}{4}$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}$

Let $P(a\cos \theta ,b\sin \theta )$ be a point on the ellipse
Equation of the normal is $ax\sec \theta -by\text{cosec}\theta =a^2-b^2$
$\therefore A(\frac{a^2-b^2}{a}\cos \theta ,0)\text{and}B(0,-\frac{a^2-b^2}{b}\sin \theta )PA=\sqrt{\frac{b^4}{a^2}{\cos }^2\theta +b^2{\sin }^2\theta }PA=\frac{b}{a}\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\text{and}PB=\sqrt{a^2{\cos }^2\theta +\frac{a^4}{b^2}{\sin }^2\theta }PB=\frac{a}{b}\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\Rightarrow PA:PB=b^2:a^2{b}^2:a^2=1:4\Rightarrow \frac{b^2}{a^2}=\frac{1}{4}\therefore e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$