If the area of the region bounded by $16 x^{2}-9 y^{2}=144$ and $8 \mathrm{x}-3 \mathrm{y}=24$ is A , then $3\left(\mathrm{~A}+6 \log _{\mathrm{e}}(3)\right)$ is equal to ______.

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Area Under Curves

If the area of the region bounded by

16 x^{2} - 9 y^{2} = 144

and

8 x - 3 y = 24

is A , then

3 (A + 6 lo g_{e} (3))

is equal to ______.

Answer: 24

The hyperbola is

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

and the line

8 x - 3 y = 24

passes through its vertex

(3, 0)

. Substituting the line into the hyperbola gives

x^{2} - 8 x + 15 = 0

, so

x = 3

and

x = 5

, between which the upper branch lies above the line. Hence

A = \int_{3}^{5} (\frac{4}{3} x^{2} - 9 - \frac{8 x - 24}{3}) d x = (\frac{40}{3} - 6 ln 3) - \frac{16}{3} = 8 - 6 ln 3.

Then

A + 6 ln 3 = 8

, so

3 (A + 6 lo g_{e} 3) = 24

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Integer answer