JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the product of the focal distances of the point \(\left(\sqrt{3}, \frac{1}{2}\right)\) on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(\mathrm{a}\gt\mathrm{b})\), be \(\frac{7}{4}\). Then the absolute difference of the eccentricities of two such ellipses is
- A \(\frac{1-\sqrt{3}}{\sqrt{2}}\)
- B \(\frac{3-2 \sqrt{2}}{2 \sqrt{3}}\)
- C \(\frac{3-2 \sqrt{2}}{3 \sqrt{2}}\)
- D \(\frac{1-2 \sqrt{2}}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(\frac{3-2 \sqrt{2}}{2 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Product of focal distances }=\left(a+e x_1\right)\left(a-e x_1\right) \\ & =a^2-e^2 x_1^2=a^2-e^2(3) \\ & =a^2-3 e^2=\frac{7}{4} \Rightarrow a^2=\frac{7}{4}+3 e^2 \\ & \Rightarrow 4 a^2=7+12 e^2 \\ & \&\left(\sqrt{3}, \frac{1}{2}\right) \text { lines on…
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