The two parabolas intersect at (0,0) and (4a,4a).
Area =∫04a(4ax−4ax2)dx
=∫04a(2ax−4ax2)dx
=[34a⋅x3/2−12ax3]04a
=332a2−316a2=316a2.
Note: Option 1 has incorrect units (a3 instead of a2); the intended answer matches 316a2.
The area bounded by the parabola y^2 = 4ax and x^2 = 4ay is :
CUET UG 2022 — Mathematics Calculus
The area bounded by the parabola y2=4ax and x2=4ay is :
Held on 7 Aug 2022 · Verified 13 Jul 2026.
Options
- A
38a3
- B
316a2
- C
332a2
- D
364a2
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