JEE Main 2022 — Hyperbola Question with Solution

By using eccentricty formula we get,

$e=\sqrt{1+\frac{b^2}{a^2}},l=\frac{2b^2}{a}$

Given ${\mathrm{e}}^2=\frac{11}{14}l$

So, $1+\frac{b^2}{a^2}=\frac{11}{14}·\frac{2b^2}{a}$

$\frac{a^2+b^2}{a^2}=\frac{11}{7}·\frac{b^2}{a} \cdots \left(1\right)$

Also $e^{\text{'}}=\sqrt{1+\frac{a^2}{{ b}^2}},l^{\text{'}}=\frac{2a^2}{ b}$

Given ${\left(e^{\text{'}}\right)}^2=\frac{11}{8}{l}^{\text{'}}$

$1+\frac{a^2}{b^2}=\frac{11}{8}·\frac{2a^2}{b}$

$\frac{a^2+b^2}{b^2}=\frac{11}{4}·\frac{a^2}{b} \cdots \left(2\right)$

Now equation $\left(1\right)÷$equation $\left(2\right)$ we get,

$\frac{b^2}{a^2}=\frac{4}{7}·\frac{b^3}{a^3}$

$\therefore 7a=4b \cdots \left(3\right)$

From $\left(2\right)$

$\frac{\frac{16b^2}{49}+b^2}{b^2}=\frac{11}{4}·\frac{16b^2}{49b}$

$\therefore b=\frac{4\times 65}{11\times 16} \cdots \left(4\right)$

We have to find value of

$77a+44b$

So, $11\left(7a+4b\right)=11\left(4b+4b\right)=11\times 8b$

$\therefore$ Value of $11\times 8b=11\times 8\times \frac{4\times 65}{16\times 11}=130$