Application of IntegralsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Area Bounded by a Hyperbola and Line: 3(A+6ln3) = 24 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
If the area of the region bounded by 16x2−9y2=14416x^2-9y^2=144 and 8x−3y=248x-3y=24 is AA, then 3(A+6log⁡e3)3(A+6\log_e 3) is equal to
Solution
Answer: 24 (± 0.01)
Step 1: The conic is a hyperbola: x29−y216=1\dfrac{x^2}{9}-\dfrac{y^2}{16}=1. From the line, 3y=8x−243y=8x-24, so y=8x−243y=\dfrac{8x-24}{3}. Step 2: Find intersections. Substitute 9y2=(8x−24)29y^2=(8x-24)^2 into 16x2−9y2=14416x^2-9y^2=144:
16x2−(8x−24)2=144 ⇒ 16x2−64(x−3)2=144 ⇒ x2−4(x−3)2=9.16x^2-(8x-24)^2=144\ \Rightarrow\ 16x^2-64(x-3)^2=144\ \Rightarrow\ x^2-4(x-3)^2=9.
 Expand: x2−4x2+24x−36=9⇒3x2−24x+45=0⇒x2−8x+15=0⇒x=3,5.x^2-4x^2+24x-36=9\Rightarrow 3x^2-24x+45=0\Rightarrow x^2-8x+15=0\Rightarrow x=3,5. Step 3: The bounded area is (upper hyperbola branch y=1316x2−144y=\tfrac13\sqrt{16x^2-144}) minus (line) from x=3x=3 to x=5x=5:
A=∫3516x2−1443 dx−12⋅2⋅163=43∫35x2−9 dx−163.A=\int_3^5\frac{\sqrt{16x^2-144}}{3}\,dx-\frac12\cdot 2\cdot\frac{16}{3}=\frac43\int_3^5\sqrt{x^2-9}\,dx-\frac{16}{3}.
 Step 4: Use ∫x2−9 dx=x2x2−9−92log⁡e(x+x2−9)\displaystyle\int\sqrt{x^2-9}\,dx=\frac{x}{2}\sqrt{x^2-9}-\frac{9}{2}\log_e\left(x+\sqrt{x^2-9}\right):
43[x2x2−9−92log⁡e(x+x2−9)]35−163.\frac43\left[\frac{x}{2}\sqrt{x^2-9}-\frac{9}{2}\log_e(x+\sqrt{x^2-9})\right]_3^5-\frac{16}{3}.
 Step 5: At x=5x=5: 25−9=4\sqrt{25-9}=4, term =52⋅4−92log⁡e9=10−92log⁡e9=\tfrac52\cdot4-\tfrac92\log_e 9=10-\tfrac92\log_e9. At x=3x=3: 0=0\sqrt{0}=0, term =0−92log⁡e3=0-\tfrac92\log_e3. Subtract:
43[10−92log⁡e9+92log⁡e3]−163.\frac43\left[10-\frac92\log_e9+\frac92\log_e3\right]-\frac{16}{3}.
 Since log⁡e9=2log⁡e3\log_e9=2\log_e3: −92(2log⁡e3)+92log⁡e3=−92log⁡e3-\tfrac92(2\log_e3)+\tfrac92\log_e3=-\tfrac92\log_e3, giving
43[10−92log⁡e3]−163=403−6log⁡e3−163=8−6log⁡e3.\frac43\left[10-\frac92\log_e3\right]-\frac{16}{3}=\frac{40}{3}-6\log_e3-\frac{16}{3}=8-6\log_e3.
 Step 6: So A=8−6log⁡e3A=8-6\log_e3, hence A+6log⁡e3=8A+6\log_e3=8 and
3(A+6log⁡e3)=3×8=24.3(A+6\log_e3)=3\times8=24.
 Correct answer: 24
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