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 :
- A \(3\)
- B \(7 / 2\)
- C \(3 / 2\)
- D \(5 / 2\)
Answer & Solution
Correct Answer
(C) \(3 / 2\)
Step-by-step Solution
Detailed explanation
\(e=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^2}{a^2}} \Rightarrow \frac{1}{2}=1-\frac{b^2}{a^2}\) \(\frac{2 b^2}{a}=14\) \(e_H=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}\) \(\left(e_H\right)^2=\frac{3}{2}\)
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