Let H _ n : x ^ 2 1 + n - y ^ 2 3 + n = 1, n ∈ ℕ . Let k be the...

Given,

Equation of hyperbola,

${H}_{n}: \dfrac{{x}^{2}}{1 + n} - \dfrac{{y}^{2}}{3 + n} = 1, n \in \mathbb{N}$

Now using the eccentricity formula we get,

${e}^{2} = 1 + \dfrac{3 + n}{1 + n}$

$⇒ {e}^{2} = \dfrac{2n + 4}{n + 1} = \dfrac{2 ( n + 2 )}{n + 1}$

Now check when $( n + 1 ) = 9, 25, 49, \ldots \ldots$ is perfect square and then at the same time $2\left(n + 2\right)$ should also be perfect square,

So checking for $n = 8 \to {e}^{2} = \dfrac{20}{9}$

$n = 24 \to {e}^{2} = \dfrac{52}{25}$

$n = 48 \to {e}^{2} = \dfrac{100}{49} ⇒ e = \dfrac{10}{7}$

Hence, for $n = 48$ both $n + 1 \text{ and } 2\left(n + 2\right)$ are perfect square,

Now we know that,

Length of latusrectum is given by $l = \dfrac{2{b}^{2}}{a} = \dfrac{2\left(n + 3\right)}{\sqrt{n + 1}} = \dfrac{2 \times 51}{7}$

Hence, $21l = \dfrac{42 \times 51}{7} = 306$