JEE Main 2026 (Online) 2nd April Evening ShiftQuestion 1
If the area of the region bounded by $16x^{2}-9y^{2}=144$ and $8x-3y=24$ is $A$, then $3(A+6\log_e(3))$ is equal to $\_\_\_\_\_\_$ .
Application of Integrals
Answer$24$
Solution
$\frac{x^2}9-\frac{y^2}{16}=1$; line $y=\frac{8x-24}3$. Substituting: $16x^2-(8x-24)^2=144\Rightarrow x^2-8x+15=0\Rightarrow x=3,5$. Between them the hyperbola arc $y=\frac43\sqrt{x^2-9}$ lies above the line (e.g. $x=4$: $3.53>2.67$). $\int_3^5\sqrt{x^2-9}\,dx=\left[\frac x2\sqrt{x^2-9}-\frac92\ln(x+\sqrt{x^2-9})\right]_3^5=10-\frac92\ln3$. $A=\frac43\left(10-\frac92\ln3\right)-\int_3^5\frac{8x-24}3dx=\frac{40}3-6\ln3-\frac{16}3=8-6\ln3$. $3(A+6\ln3)=3\cdot8=24$.