JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(\mathrm{E}: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}\gt\mathrm{b}\) and \(\mathrm{H}: \frac{x^2}{\mathrm{~A}^2}-\frac{y^2}{\mathrm{~B}^2}=1\). Let the distance between the foci of E and the foci of \(H\) be \(2 \sqrt{3}\). If \(a-A=2\), and the ratio of the eccentricities of \(E\) and \(H\) is \(\frac{1}{3}\), then the sum of the lengths of their latus rectums is equal to:
- A 10
- B 9
- C 8
- D 7
Answer & Solution
Correct Answer
(C) 8
Step-by-step Solution
Detailed explanation
\begin{aligned} & \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \text { foci are }(\mathrm{ae}, 0) \text { and }(-\mathrm{ae}, 0) \\ & \frac{\mathrm{x}^2}{\mathrm{~A}^2}+\frac{\mathrm{y}^2}{\mathrm{~B}^2}=1 \text { foci are }\left(\mathrm{Ae}^{\prime},…
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