### Ellipse Question 4

#### Question 4 - 31 January - Shift 1

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:

(1) $\frac{1}{\sqrt{2}}$

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

(3) $\frac{1}{2}$

(4) $\frac{\sqrt{3}}{4}$

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Equation of Normal ( Parametric form ), Distance between point and line

Equation of normal is

$2 x \sec \theta-b\cdot y cosec \theta=4-b^{2}$

Distance from $(0,0)=\frac{4-b^{2}}{\sqrt{4 \sec ^{2} \theta+b^{2} cosec^{2} \theta}}$

Distance is maximum if $4 \sec ^{2} \theta+b^{2} cosec^{2} \theta$ 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 $

$\Rightarrow e=\frac{\sqrt{3}}{2}$

$\therefore$ The eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$.