Hyperbola
Eccentricity of Conjugate/Related Hyperbola
nta_pyq_2024_jan
Grade 11
Question:
Let $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a>b$ be an ellipse, whose eccentricity is $\dfrac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is:
Step-by-Step Solution
Key Concept: From ellipse eccentricity $e=1/\sqrt2$: $b^2/a^2=1/2$. From LR $=2b^2/a=\sqrt{14}$: find $a$ and $b^2$. Then compute $e_H^2=1+b^2/a^2$ for the hyperbola.
$e_{ellipse}=1/\sqrt2\Rightarrow b^2/a^2=1/2$. $2b^2/a=\sqrt{14}$: $a=\sqrt7$, $b^2=7/2$. Hyperbola: $e_H^2=1+b^2/a^2=1+1/2=3/2$.
Correct Answer: 3