Let H _{ n }=frac{ x ^2}{1+ n }-frac{ y ^2}{3 | vedclass

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Question

$Hn \Rightarrow \frac{ x ^2}{1+ n }-\frac{ y ^2}{3+ n }=1$

$e =\sqrt{1+\frac{ b ^2}{ a ^2}}=\sqrt{1+\frac{3+ n }{1+ n }}=\sqrt{\frac{2 n +4}{ n +1}

Vedclass · Accepted Answer

$306$

Vedclass · Answer

$Hn \Rightarrow \frac{ x ^2}{1+ n }-\frac{ y ^2}{3+ n }=1$

$e =\sqrt{1+\frac{ b ^2}{ a ^2}}=\sqrt{1+\frac{3+ n }{1+ n }}=\sqrt{\frac{2 n +4}{ n +1}}$

$e =\sqrt{\frac{2 n +4}{ n +1}}$

$n =48(	ext { smallest even value for which } e \in Q )$

$e =\frac{10}{7}$

$a ^2 =n+1 \quad b ^2=n+3$

$=49 \quad, \quad=51$

$1 =	ext { length of } LR =\frac{2 b ^2}{ a }$

$L =2 \cdot \frac{51}{7}$

$1 =\frac{102}{7}$

$21 \ell=306$

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 $.......$.

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