Let Let be the smallest even value of such that the eccentricity of is a rational number. If is the length of the latus rectum of then is equal to _________. [2023]

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Solution

Q.
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 the length of the latus rectum of $H_{k},$ then $21 l$ is equal to _________. [2023]

Ans.
(306)

$H_{n} = \frac{x^{2}}{1 + n} - \frac{y^{2}}{3 + n} = 1 be a hyperbola$

$Eccentricity of hyperbola, e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{3 + n}{1 + n}} = \sqrt{\frac{2 n + 4}{n + 1}}$

$For rational eccentricity, \sqrt{\frac{2 n + 4}{n + 1}} should be a perfect square.$

$∴$ $n = 48 (smallest even value for which e \in Q)$

$∴$ $e = \frac{10}{7}$

$Now, a^{2} = n + 1 = 49, b^{2} = n + 3 = 51$

$l = length of latus rectum = \frac{2 b^{2}}{a}$

$\Rightarrow l = 2 \cdot \frac{51}{7} = \frac{102}{7}$ $∴$ $21 l = 306$

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