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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:

  1. A 10
  2. B 9
  3. C 8
  4. D 7
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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},…