JEE MAIN - Mathematics (2023 - 11th April Morning Shift - No. 18)
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 ____________.
Answer
306
Explanation
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 $b^2=3+n$
$$ \begin{aligned} \operatorname{Eccentricity}(e) & =\sqrt{1+\frac{b^2}{a^2}} \\\\ & =\sqrt{1+\left(\frac{3+n}{1+n}\right)}=\sqrt{\frac{2 n+4}{n+1}}=\sqrt{\frac{2(n+2)}{n+1}} \end{aligned} $$
The smallest even value for which $e \in Q$ is 48 .
$$ \begin{aligned} \therefore n & =48 \\\\ \therefore e & =\sqrt{\frac{2(48+2)}{48+1}}=\frac{10}{7} \end{aligned} $$
$$ \begin{array}{ll} \Rightarrow a^2=n+1=49, b^2=n+3=51 \\\\ \therefore 21 l=21 \times\left(\frac{2 b^2}{a}\right)=21 \times 2 \times \frac{51}{7}=306 \end{array} $$
$$ H_n \Rightarrow \frac{x^2}{1+n}-\frac{y^2}{3+n}=1, n \in N $$
Here, $a^2=1+n$ and $b^2=3+n$
$$ \begin{aligned} \operatorname{Eccentricity}(e) & =\sqrt{1+\frac{b^2}{a^2}} \\\\ & =\sqrt{1+\left(\frac{3+n}{1+n}\right)}=\sqrt{\frac{2 n+4}{n+1}}=\sqrt{\frac{2(n+2)}{n+1}} \end{aligned} $$
The smallest even value for which $e \in Q$ is 48 .
$$ \begin{aligned} \therefore n & =48 \\\\ \therefore e & =\sqrt{\frac{2(48+2)}{48+1}}=\frac{10}{7} \end{aligned} $$
$$ \begin{array}{ll} \Rightarrow a^2=n+1=49, b^2=n+3=51 \\\\ \therefore 21 l=21 \times\left(\frac{2 b^2}{a}\right)=21 \times 2 \times \frac{51}{7}=306 \end{array} $$
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