JEE Main 2023 — Hyperbola Question with Solution

The given equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,ae=2,e=\frac{3}{2}$

$\Rightarrow a=\frac{4}{3}$

$\Rightarrow b^2=a^2{e}^2-a^2=4-\frac{16}{9}=\frac{20}{9}$

The tangent is perpendicular to $2x+3y=6$.

Slope of tangent is $m=\frac{-1}{{\displaystyle \frac{-2}{3}}}=\frac{3}{2}$

Equation of tangent, $m=\frac{3}{2}$

Equation of tangent is $y=mx-\sqrt{a^2{m}^2-b^2}$
$\Rightarrow y=\frac{3}{2}x-\sqrt{\frac{16}{9}\times \frac{9}{4}-\frac{20}{9}}$

( $\because$ Tangent is in $1^{\text{st }}$ quadrant$\left.\Rightarrow C<0\right)$

$\Rightarrow y=\frac{3}{2}x-\frac{4}{3}$

Converting the equation into intercept form:

$\Rightarrow \frac{x}{{\displaystyle \left(\frac{8}{9}\right)}}+\frac{y}{\left(-{\displaystyle \frac{4}{3}}\right)}=1$

$\Rightarrow \left|6a\right|+\left|5b\right|=\left|\frac{6\times 8}{9}\right|+\left|\frac{5\times \left(-4\right)}{3}\right|=12$

Hence this is the required option.