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

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(B) The equation of the hyperbola is $H_{n} \Rightarrow \frac{x^2}{1+n} - \frac{y^2}{3+n} = 1$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b

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$306$

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(B) The equation of the hyperbola is $H_{n} \Rightarrow \frac{x^2}{1+n} - \frac{y^2}{3+n} = 1$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{3+n}{1+n}} = \sqrt{\frac{1+n+3+n}{1+n}} = \sqrt{\frac{2n+4}{n+1}}$.For $e$ to be a rational number,$\frac{2n+4}{n+1}$ must be a perfect square of a rational number. Let $\frac{2n+4}{n+1} = q^2$.We test even values of $n$: If $n=2$,$e = \sqrt{8/3}$ (not rational). If $n=4$,$e = \sqrt{12/5}$ (not rational). If $n=6$,$e = \sqrt{16/7}$ (not rational). If $n=8$,$e = \sqrt{20/9}$ (not rational).We need $2n+4 = m^2(n+1)$. For $n=48$,$e = \sqrt{\frac{2(48)+4}{48+1}} = \sqrt{\frac{100}{49}} = \frac{10}{7}$,which is rational.Thus,$k = 48$. The length of the latus rectum $l = \frac{2b^2}{a} = \frac{2(k+3)}{\sqrt{k+1}} = \frac{2(48+3)}{\sqrt{48+1}} = \frac{2(51)}{7} = \frac{102}{7}$.Therefore,$21l = 21 	imes \frac{102}{7} = 3 	imes 102 = 306$.

Let $H_{n} = \frac{x^2}{1+n} - \frac{y^2}{3+n} = 1$,where $n \in N$. Let $k$ be the smallest even value of $n$ such that the eccentricity of $H_{k}$ is a rational number. If $l$ is the length of the latus rectum of $H_{k}$,then $21l$ is equal to $.......$.

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