JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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 \(\frac{\sqrt{3}}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{\sqrt{3}}{2}\)
Step-by-step Solution
Detailed explanation
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…
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