JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the foci of hyperbola coincide with the foci of the ellipse \( \frac{x^{2}}{36}+\frac{y^{2}}{16}=1. \) If the eccentricity of the hyperbola is 5, then the length of its latus rectum is:
- A 12
- B 16
- C \( \frac{96}{\sqrt{5}} \)
- D \( 24\sqrt{5} \)
Answer & Solution
Correct Answer
(C) \( \frac{96}{\sqrt{5}} \)
Step-by-step Solution
Detailed explanation
Let \( e_1 \) be eccentricity of ellipse \( \Rightarrow e_{1}=\sqrt{1-\frac{16}{36}}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3} \) So \( ae_{1}=6.\frac{\sqrt{5}}{3}=2\sqrt{5} \) Now H: \( \frac{x^{2}}{p^{2}}-\frac{y^{2}}{q^{2}}=1 \) \( p.e=ae_{1}\) \(p.5=2\sqrt{5}\)…
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