JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The eccentricity of an ellipse \(E\) with centre at the origin \(O\) is \(\dfrac{\sqrt{3}}{2}\) and its directrices are \(x = \pm \dfrac{4\sqrt{6}}{3}\). Let \(H: \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) be a hyperbola whose eccentricity is equal to the length of semi-major axis of \(E\), and whose length of latus rectum is equal to the length of minor axis of \(E\). Then the distance between the foci of \(H\) is :
- A \(\dfrac{4\sqrt{2}}{\sqrt{7}}\)
- B \(\dfrac{4\sqrt{2}}{7}\)
- C \(\dfrac{4}{\sqrt{7}}\)
- D \(\dfrac{8}{7}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{8}{7}\)
Step-by-step Solution
Detailed explanation
For the ellipse \(E\), the eccentricity is \(e_E = \dfrac{\sqrt{3}}{2}\) and the directrices are \(x = \pm \dfrac{a_E}{e_E} = \pm \dfrac{4\sqrt{6}}{3}\). The semi-major axis \(a_E\) is given by:…
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