JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1
The normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$ at the point $(8,3\sqrt{3})$ on it passes through the point :
Hyperbola
Solution
$(8,3\sqrt3)$ on the curve: $\dfrac{64}{a^2}-3=1\Rightarrow a^2=16$. Normal: $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2$: $2x+\sqrt3y=25$. For $(-1,9\sqrt3)$: $-2+27=25$ ✓.