Hyperbola Question 2
Question 2 - 2024 (01 Feb Shift 1)
Let $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1, \mathrm{a}>\mathrm{b}$ 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
$e=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^{2}}{a^{2}}} \Rightarrow \frac{1}{2}=1-\frac{b^{2}}{a^{2}}$
$\frac{2 b^{2}}{a}=14$
$e_{H}=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$
$\left(e_{H}\right)^{2}=\frac{3}{2}$
