Let H_n:x^21+n-y^23+n=1,n N.

Question

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 $21l$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 306

Accepted Answer

We have,

${H}_{n}\Rightarrow \frac{{x}^{2}}{1+n}-\frac{{y}^{2}}{3+n}=1,n\in N$

Here, ${a}^{2}=1+n$ and ${b}^{2}=3+n$

$\begin{gathered} Eccentricity(e) =\sqrt{1+\frac{{b}^{2}}{{a}^{2}}} \ \ =\sqrt{1+(\frac{3+n}{1+n})}=\sqrt{\frac{2n+4}{n+1}}=\sqrt{\frac{2(n+2)}{n+1}} $

The smallest even value for which $e\in Q$ is 48 .

$ 	herefore n =48 \ \ 	herefore e =\sqrt{\frac{2(48+2)}{48+1}}=\frac{10}{7} \end{gathered}$

$\begin{gathered} \Rightarrow {a}^{2}=n+1=49,{b}^{2}=n+3=51 \ 	herefore 21l=21	imes (\frac{2{b}^{2}}{a})=21	imes 2	imes \frac{51}{7}=306 \end{gathered}$

Solutions · 1