Let $H_n:\frac{x^2}{1+n}-\frac{y^2}{3+n}=1, n\in \mathrm{\mathbb{N}}$. 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

Given,

Equation of hyperbola,

$H_n:\frac{x^2}{1+n}-\frac{y^2}{3+n}=1, n\in \mathrm{\mathbb{N}}$

Now using the eccentricity formula we get,

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

$\Rightarrow e^2=\frac{2n+4}{n+1}=\frac{2(n+2)}{n+1}$

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

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

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

$n=48\to e^2=\frac{100}{49}\Rightarrow e=\frac{10}{7}$

Hence, for $n=48$ both $n+1 \& 2\left(n+2\right)$ are perfect square,

Now we know that,

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

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