Coordinate Geometry
Eccentricity of hyperbola; latus rectum
MMTS_Full_Test_21
Grade 12
Question:
Let $H_n: \dfrac{x^2}{1+n}-\dfrac{y^2}{3+n}=1$, $n\in\mathbb{N}$. Let $k$ be the smallest even value of $n$ such that eccentricity of $H_k$ is rational. If $\ell$ is the length of the latus rectum of $H_k$, then $21\ell$ is equal to
(A) 101
(B) 204
(C) 102
(D) 306
Step-by-Step Solution
Key Concept: Eccentricity: $e=\sqrt{1+\frac{3+n}{1+n}}=\sqrt{\frac{2n+4}{n+1}}$. For $e$ to be rational, $\frac{2n+4}{n+1}$ must be a perfect square of a rational number.
Smallest even $n=48$, $e=10/7$. $\ell=102/7$, $21\ell=306$.
Correct Answer: (D) 306