The area (in sq. units) bounded by the parabola \(y^2 = 4ax\), its latus rectum and the \(x\)-axis in the first quadrant is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{4}{3}a^2\)
The focus of \(y^2 = 4ax\) is (a, 0), so the latus rectum is the line x = a. In the first quadrant, \(y = 2\sqrt{a}\,\sqrt{x}\), and the region runs from x = 0 to x = a:
\( \displaystyle \text{Area} = \int_{0}^{a} 2\sqrt{a}\,x^{1/2}\,dx \)
\( \displaystyle = 2\sqrt{a}\left[\frac{2}{3}x^{3/2}\right]_{0}^{a} \)
\( \displaystyle = 2\sqrt{a} \times \frac{2}{3}a^{3/2} = \frac{4}{3}a^2 \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{3}a^2\) — Incorrect. The area is \(\frac{4}{3}a^2\), four times this value.
Option (2) \(\displaystyle\frac{2}{3}a^2\) — Incorrect. The area is \(\frac{4}{3}a^2\), twice this value.
Option (4) \(\displaystyle\frac{8}{3}a^2\) — Incorrect. This is the area enclosed by the parabola and the whole latus rectum, above and below the x-axis. The question asks only for the first-quadrant half.