ExamBro
ExamBro
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:

  1. A 12
  2. B 16
  3. C \( \frac{96}{\sqrt{5}} \)
  4. D \( 24\sqrt{5} \)
Verified Solution

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: \( \frac{x^{2}}{p^{2}}-\frac{y^{2}}{q^{2}}=1 \) \( p.e=ae_{1}\) \(p.5=2\sqrt{5}\)…
Same subject
Explore more questions on app