Hyperbola Question 2
Question 2 - 2024 (01 Feb Shift 1)
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) 3
(2) $7 / 2$
(3) $3 / 2$
(4) $5 / 2$
Show Answer
Answer (3)
Solution
$e=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^{2}}{a^{2}}} \Rightarrow \frac{1}{2}=1-\frac{b^{2}}{a^{2}}$
$\frac{2 b^{2}}{a}=14$
$e _H=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$
$\left(e _H\right)^{2}=\frac{3}{2}$
