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STD 11 - 10.2 parabola,ellipse,hyperbola — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 10.2 parabola,ellipse,hyperbola4 Marks
MCQ
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
    $7 / 2$
  • ✓
     $3 / 2$
  • D
    $5 / 2$
✓

Answer

Correct option: C.
 $3 / 2$
c
$e=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^2}{a^2}} \Rightarrow \frac{1}{2}=1-\frac{b^2}{a^2}$

$\frac{2 b^2}{a}=14$

$e_H=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$

$\left(e_H\right)^2=\frac{3}{2}$

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