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Mathematics · Area under the Curves

JEE Main 2025 — 22 January, Evening Shift — Question 20

The area of the region enclosed by the curves y=x2−4x+4y = x^2 - 4x + 4 and y2=16−8xy^2 = 16 - 8x is:

  1. Option A:

    83\frac{8}{3}

    Correct
  2. Option B:

    43\frac{4}{3}

  3. Option C:

    55

  4. Option D:

    88

Answer: A

Step-by-step solution

y=(x−2)2,y2=−8(x−2)y = (x-2)^2, \quad y^2 =- 8(x-2) y=x2,y2=−8xy = x^2, \quad y^2 = -8x =16ab3=16×14×23=83= \frac{16ab}{3} = \frac{16 \times \frac{1}{4} \times 2}{3} = \frac{8}{3}
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
22 January, Evening Shift
Subject
Mathematics
Chapter
Area under the Curves
Topic
Area under the Curves
The area of the region enclosed by the curves y = x 2 - 4x + 4 and y… | JEE Main 2025 PYQ with Solution · DhiX AI