JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Hyperbola

With foci $(\pm 2,0)$ and $e=\frac{3}{2}$, the hyperbola $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$ has $Ae=2$, so $A=\frac{4}{3}$ and $B^2=A^2(e^2-1)=4-\frac{16}{9}=\frac{20}{9}$. The line $2x+3y=6$ has slope $-\frac{2}{3}$, so the required tangent has slope $m=\frac{3}{2}$. A tangent of slope $m$ is $y=mx+c$ with $c^2=A^2{m}^2-B^2=4-\frac{20}{9}=\frac{16}{9}$, hence $c=\pm \frac{4}{3}$. The point of contact $\left(-\frac{A^{2}m}{c},-\frac{B^2}{c}\right)$ lies in the first quadrant only when $c=-\frac{4}{3}$, giving the tangent $y=\frac{3}{2}x-\frac{4}{3}$, which touches at $\left(2,\frac{5}{3}\right)$. Its $x$-intercept is $a=\frac{8}{9}$ and its $y$-intercept is $b=-\frac{4}{3}$, so $|6a|+|5b|=6\cdot \frac{8}{9}+5\cdot \frac{4}{3}=\frac{48}{9}+\frac{60}{9}=12$.