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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

Let \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b\) be an ellipse, whose eccentricity is \(\frac{1}{\sqrt{2}}\) and the length of the latus rectum is \(\sqrt{14}\). Then the square of the eccentricity of \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is :

  1. A \(3\)
  2. B \(7 / 2\)
  3. C  \(3 / 2\)
  4. D \(5 / 2\)
Verified Solution

Answer & Solution

Correct Answer

(C)  \(3 / 2\)

Step-by-step Solution

Detailed explanation

\(e=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^2}{a^2}} \Rightarrow \frac{1}{2}=1-\frac{b^2}{a^2}\) \(\frac{2 b^2}{a}=14\) \(e_H=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}\) \(\left(e_H\right)^2=\frac{3}{2}\)