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10-2. Parabola, Ellipse, Hyperbola
hard
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
$\frac{1}{\sqrt{2}}$
B
$\frac{\sqrt{3}}{2}$
C
$\frac{1}{2}$
D
$\frac{\sqrt{3}}{4}$
(JEE MAIN-2023)
Solution
Equation of normal is
$2 x \sec \theta-b y \operatorname{cosec} \theta=4-b^2$
Distance from $(0,0)=\frac{4-b^2}{\sqrt{4 \sec ^2 \theta+b^2 \operatorname{cosec}^2 \theta}}$
Distance is maximum if
$4 \sec ^2 \theta+b^2 \operatorname{cosec}^2 \theta$ is minimum
$\Rightarrow \tan ^2 \theta=\frac{b}{2}$
$\Rightarrow \frac{4-b^2}{\sqrt{4 \cdot \frac{b+2}{2}+b^2 \cdot \frac{b+2}{b}}}=1$
$\Rightarrow 4-b^2=b+2 \Rightarrow b=1 \Rightarrow e =\frac{\sqrt{3}}{2}$
Standard 11
Mathematics
