JEE Main 2026 — Area Under Curves Question with Solution

The given equations are:

Hyperbola: $16x^2-9y^2=144\Rightarrow \frac{x^2}{9}-\frac{y^2}{16}=1$

Line: $8x-3y=24\Rightarrow y=\frac{8}{3}(x-3)$

To find the points of intersection, substitute $y$ from the line equation into the hyperbola equation:

$16x^2-9{\left(\frac{8}{3}(x-3)\right)}^2=144$

$16x^2-64(x^2-6x+9)=144$

$x^2-4(x^2-6x+9)=9$

$-3x^2+24x-45=0$

$x^2-8x+15=0$

$(x-3)(x-5)=0\Rightarrow x=3,x=5$

For $x\in [3,5]$, the upper curve is the hyperbola $y=\frac{4}{3}\sqrt{x^2-9}$ and the lower curve is the line $y=\frac{8}{3}(x-3)$.

The area $A$ of the bounded region is:

$A={\int }_3^5\left(\frac{4}{3}\sqrt{x^2-9}-\frac{8}{3}(x-3)\right)dx$

Using the standard integral formula $\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln |x+\sqrt{x^2-a^2}|$:

${\int }_3^5\sqrt{x^2-9}dx={\left[\frac{x}{2}\sqrt{x^2-9}-\frac{9}{2}\ln |x+\sqrt{x^2-9}|\right]}_3^5$

$=\left(\frac{5}{2}(4)-\frac{9}{2}\ln 9\right)-\left(0-\frac{9}{2}\ln 3\right)$

$=10-9\ln 3+\frac{9}{2}\ln 3=10-\frac{9}{2}\ln 3$

For the linear part:

${\int }_3^5(x-3)dx={\left[\frac{(x-3{)}^2}{2}\right]}_3^5=\frac{4}{2}-0=2$

Substituting these back into the area equation:

$A=\frac{4}{3}\left(10-\frac{9}{2}\ln 3\right)-\frac{8}{3}(2)$

$A=\frac{40}{3}-6\ln 3-\frac{16}{3}$

$A=8-6\ln 3$

We need to find the value of $3(A+6{\log }_{e}3)$:

$A+6\ln 3=8$

$3(A+6\ln 3)=3\times 8=24$

Answer: $24$