Hyperbola
Eccentricity Relation Between Hyperbola and Ellipse
nta_pyq_2024_jan
Grade 11
Question:
For $0<\theta<\pi/2$, if the eccentricity of the hyperbola $x^2-y^2\csc^2\theta=5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^2\csc^2\theta+y^2=5$, then the value of $\theta$ is:
$\dfrac{\pi}{6}$
$\dfrac{5\pi}{12}$
$\dfrac{\pi}{3}$
$\dfrac{\pi}{4}$
Step-by-Step Solution
Key Concept: Write the hyperbola as $\frac{x^2}{5}-\frac{y^2}{5\sin^2\theta}=1$ and the ellipse as $\frac{x^2}{5\sin^2\theta}+\frac{y^2}{5}=1$. Compute $e_h$ and $e_c$ in terms of $\sin\theta$, then use $e_h=\sqrt{7}e_c$.
$e_h=\sqrt{1+\sin^2\theta}$, $e_c=\sqrt{1-\sin^2\theta}$. $e_h=\sqrt7e_c\Rightarrow1+\sin^2\theta=7-7\sin^2\theta\Rightarrow\sin^2\theta=3/4\Rightarrow\theta=\pi/3$.
Correct Answer: 3