JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 24

Intersections. The hyperbola $\frac{x^2}9-\frac{y^2}{16}=1$ meets $y=\frac{8x-24}3$ where $x^2-8x+15=0$: at $(3,0)$ and $\left(5,\frac{16}3\right)$. Between them the hyperbola lies above the line.

Area.
$$A=\int_3^5\left[\frac43\sqrt{x^2-9}-\frac{8x-24}3\right]dx=\frac43\left(10-\frac92\ln3\right)-\frac{16}3=8-6\ln3,$$
using $\int\sqrt{x^2-9}\,dx=\frac x2\sqrt{x^2-9}-\frac92\ln\left|x+\sqrt{x^2-9}\right|$.

Value. $3(A+6\ln3)=24$.

Answer: 24