JEE Main 2023 — Hyperbola Question with Solution

Given that the foci are at $(\pm 2,0)$, the distance between the foci is $2ae=2\times 2=4$.

So, $a=\frac{4}{3}$.

The eccentricity of the hyperbola is given as $\frac{3}{2}$. Thus, $e=\frac{3}{2}$.

For a hyperbola, $e^2=1+\frac{b^2}{a^2}$, which gives $b^2=a^2(e^2-1)$.

Substituting the given values, we get $b^2=\frac{16}{9}-\frac{16}{9}=\frac{20}{9}$.

The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, or $\frac{9x^2}{16}-\frac{9y^2}{20}=1$.

The slope of the tangent line, which is perpendicular to the given line $2x+3y=6$, is $\frac{3}{2}$.

The equation of the tangent to a hyperbola is $y=mx\pm \sqrt{a^2{m}^2-b^2}$. Substituting the given values, we get $y=\frac{3x}{2}\pm \sqrt{\frac{16}{9}\times \frac{9}{4}-\frac{20}{9}}=\frac{3x}{2}\pm \frac{4}{3}$.

The $x$-intercept occurs when $y=0$, so $|a|=\frac{8}{9}$, and the $y$-intercept occurs when $x=0$, so $|b|=\frac{4}{3}$.

Finally, $|6a|+|5b|=6\frac{8}{9}+5\frac{4}{3}=\frac{48}{9}+\frac{60}{9}=\frac{108}{9}=12$.