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QJEEM23Maf5d1: JEE Main 2023 Mathematics - Class 11 JEEM23Maf5d1 Solution: Let . Let be the smallest even value of such that the…

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

Step-by-Step Verified Solution

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{aligned}\operatorname{Eccentricity}(e) &amp; =\sqrt{1+\frac{b^2}{a^2}} \\\\&amp; =\sqrt{1+\left(\frac{3+n}{1+n}\right)}=\sqrt{\frac{2 n+4}{n+1}}=\sqrt{\frac{2(n+2)}{n+1}}\end{aligned}

The smallest even value for which

e \in Q

is 48 .

\begin{aligned}\therefore n &amp; =48 \\\\\therefore e &amp; =\sqrt{\frac{2(48+2)}{48+1}}=\frac{10}{7}\end{aligned}

\begin{array}{ll}\Rightarrow a^2=n+1=49, b^2=n+3=51 \\\\\therefore 21 l=21 \times\left(\frac{2 b^2}{a}\right)=21 \times 2 \times \frac{51}{7}=306\end{array}

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