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Find the area enclosed by the parabola y² = 4a²(x - 1), the vertical line x = 1, and the horizontal line y = 4a:
- 16a/3 sq unit
- 5a sq unit
- 17a/4 sq unit
- None of these
Correct answer: 16a/3 sq unit
Solution
On the parabola x-1=y^2/(4a^2). Area = Integral_0^{4a} y^2/(4a^2) dy = (1/(4a^2))*(4a)^3/3 = 16a/3 sq unit.
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