JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Hyperbola

From $e=\frac{5}{4}$, $e^2=1+\frac{b^2}{a^2}=\frac{25}{16}$ gives $\frac{b^2}{a^2}=\frac{9}{16}$. Requiring that $(\frac{8}{\sqrt{5}},\frac{12}{5})$ lie on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and using this ratio gives $a^2=\frac{64}{25}$ and $b^2=\frac{36}{25}$. The normal to the hyperbola at $(x_0,y_0)$ is $\frac{a^{2}x}{x_0}+\frac{b^{2}y}{y_0}=a^2+b^2$, and here $a^2+b^2=\frac{100}{25}=4$. Substituting $x_0=\frac{8}{\sqrt{5}}$ and $y_0=\frac{12}{5}$ gives $\frac{8\sqrt{5}}{25}x+\frac{15}{25}y=4$; clearing the denominator yields $8\sqrt{5}x+15y=100$. Matching this with $8\sqrt{5}x+\beta y=\lambda$ identifies $\beta =15$ and $\lambda =100$, so $\lambda -\beta =85$.