Ellipse
Maximum distance of normal from origin — eccentricity
nta_pyq_2023_jan
Grade None
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:
(1) $\frac{1}{\sqrt{2}}$
(2) $\frac{\sqrt{3}}{2}$
(3) $\frac{1}{2}$
(4) $\frac{\sqrt{3}}{4}$
\frac{1}{\sqrt{2}}
\frac{\sqrt{3}}{2}
\frac{1}{2}
\frac{\sqrt{3}}{4}
Step-by-Step Solution
Key Concept: Normal to $\frac{x^2}{4}+\frac{y^2}{b^2}=1$ at point: $2x\sec\theta - by\csc\theta = 4 - b^2$. Distance from origin $= \frac{4-b^2}{\sqrt{4\sec^2\theta + b^2\csc^2\theta}}$. Maximize by minimizing denominator; set $\tan^2\theta = b/2$, then equate max distance to 1.
Normal: $2x\sec\theta - by\csc\theta = 4-b^2$. Distance from origin $= \frac{4-b^2}{\sqrt{4\sec^2\theta+b^2\csc^2\theta}}$. Minimize denominator: $\tan^2\theta = b/2$, denominator $= \sqrt{(b+2)^2/b \cdot b} \cdot$ (simplify) $= \sqrt{(b+2)^2} \cdot$ correction term. Setting max distance $=1$: $4-b^2 = b+2 \Rightarrow b=1$. $a=2$, $e = \sqrt{1-1/4} = \frac{\sqrt{3}}{2}$.
Correct Answer: 2