JEE Main 2026 — 2 April, Evening Shift — Question 46

$16 x^{2}-9 y^{2}=144$ $16 x^{2}-(8 x-24)^{2}=144$ $16 x^{2}-64(x-3)^{2}=144 \Rightarrow x^{2}-4(x-3)^{2}=9$ $3 \mathrm{x}^{2}-24 \mathrm{x}+45 \Rightarrow \mathrm{x}^{2}-8 \mathrm{x}+15=0$ $\mathrm{x}=3,5$ Area $=\int_{3}^{5} \sqrt{\frac{16 \mathrm{x}^{2}-144}{9}}-\frac{1}{2} \cdot 2 \cdot \frac{16}{3}$ $=\frac{4}{3} \int_{3}^{5} \sqrt{x^{2}-9}-\frac{16}{3}$ $=\frac{4}{3}\left(\frac{x}{2} \sqrt{x^{2}-9}-\frac{9}{2} \log _{e}\left(x+\sqrt{x^{2}-9}\right)\right)_{3}^{5}-\frac{16}{3}$ $=\frac{4}{3}\left(\frac{5}{2} \cdot 4-\frac{9}{2} \log _{e} 9-\frac{3}{2} \cdot 0+\frac{9}{2} \log _{e} 3\right)-\frac{16}{3}$ Area $=8-6 \log _{e} 3=A$ $\therefore \mathrm{A}+6 \ell \mathrm{n} 3=8$ $\Rightarrow 3(\mathrm{~A}+6 \ell \mathrm{n} 3)=24$