JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Hyperbola

The hyperbola $3x^2-4y^2=36$ is $\frac{x^2}{12}-\frac{y^2}{9}=1$. The point nearest the line $3x+2y=1$ is where the tangent is parallel to it, so the tangent slope equals $-\frac{3}{2}$. The tangent at $(x_0,y_0)$ is $\frac{x{x}_0}{12}-\frac{y{y}_0}{9}=1$ with slope $\frac{3x_0}{4y_0}=-\frac{3}{2}$, giving $x_0=-2y_0$. Substituting into the hyperbola, $3(4y_0^2)-4y_0^2=36$ gives $y_0^2=\frac{9}{2}$, so $y_0=\pm \frac{3}{\sqrt{2}}$. The branch closer to the line is $\left(\frac{6}{\sqrt{2}},-\frac{3}{\sqrt{2}}\right)$, so $\sqrt{2}(y_0-x_0)=\sqrt{2}\left(-\frac{3}{\sqrt{2}}-\frac{6}{\sqrt{2}}\right)=-9$.