The area (in sq. units) bounded by a parabola \( y^2=4x \) and its latus rectum, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{8}{3} \)
For \( y^2=4ax \) with a = 1, the latus rectum is the line x = 1. The region is symmetric about the x-axis, so the area is twice the part above it:
\( \displaystyle \text{Area}=2\int_0^1 2\sqrt{x}\,dx \)
\( \displaystyle =4\left[\frac{2}{3}x^{3/2}\right]_0^1 \)
\( \displaystyle =\frac{8}{3} \)
Why the other options are incorrect
Option (2) \( \displaystyle\frac{4}{3} \) — Incorrect. This is only the half above the x-axis; the region lies on both sides of it.
Option (3) \( \displaystyle\frac{2}{3} \) — Incorrect. The area is \( \frac{8}{3} \), not \( \frac{2}{3} \).
Option (4) \( \displaystyle\frac{16}{3} \) — Incorrect. The area is \( \frac{8}{3} \), not \( \frac{16}{3} \).