Question

If the maximum distance of normal to the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1,b<2$, from the origin is 1 , then the eccentricity of the ellipse is :

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

Answer 1

Answer: (B)

Equation of normal is
$2x\sec \theta -\text{ by }\mathrm{cosec}\theta =4-b^2$
Distance from $(0,0)=\frac{4-b^2}{\sqrt{4{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta }}$
Distance is maximum if
$4{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta$ is minimum
$\Rightarrow {\tan }^2\theta =\frac{b}{2}$
$\Rightarrow \frac{4-b^2}{\sqrt{4\cdot \frac{b+2}{2}+b^2\cdot \frac{b+2}{b}}}=1$
$\Rightarrow 4-b^2=b+2\Rightarrow b=1\Rightarrow e=\frac{\sqrt{3}}{2}$