If the maximum distance of normal to the ellipse (\frac{x^2}... | Mathematics | SolveeIt | Solveeit

Today, August 18

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

$\frac{\sqrt{3}}{2}$

Explanation

Solution

Equation of normal is
2xsecθ− by cosecθ=4−b2
Distance from (0,0)=4sec2θ+b2cosec2θ4−b2
Distance is maximum if
4sec2θ+b2cosec2θ is minimum
⇒tan2θ=2b
⇒4⋅2b+2+b2⋅bb+24−b2=1
⇒4−b2=b+2⇒b=1⇒e=23