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10-2. Parabola, Ellipse, Hyperbola
hard
Let $H _{ n }=\frac{ x ^2}{1+ n }-\frac{ y ^2}{3+ n }=1, n \in N$. Let $k$ be the smallest even value of $n$ such that the eccentricity of $H _{ k }$ is a rational number. If $l$ is length of the latus return of $H _{ k }$, then $21 l$ is equal to $.......$.
A
$305$
B
$306$
C
$304$
D
$303$
(JEE MAIN-2023)
Solution
$Hn \Rightarrow \frac{ x ^2}{1+ n }-\frac{ y ^2}{3+ n }=1$
$e =\sqrt{1+\frac{ b ^2}{ a ^2}}=\sqrt{1+\frac{3+ n }{1+ n }}=\sqrt{\frac{2 n +4}{ n +1}}$
$e =\sqrt{\frac{2 n +4}{ n +1}}$
$n =48(\text { smallest even value for which } e \in Q )$
$e =\frac{10}{7}$
$a ^2 =n+1 \quad b ^2=n+3$
$=49 \quad, \quad=51$
$1 =\text { length of } LR =\frac{2 b ^2}{ a }$
$L =2 \cdot \frac{51}{7}$
$1 =\frac{102}{7}$
$21 \ell=306$
Standard 11
Mathematics
