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