The area of region bounded by the curve \(y^2 = 4ax\) and the straight line \(x = 2a, a \gt 0\) in the first quadrant is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{8\sqrt{2}a^2}{3}\) sq. units
In the first quadrant the parabola is \(y = 2\sqrt{a}\,\sqrt{x}\). The region lies under this arc from x = 0 to the line x = 2a:
\(\displaystyle A = \int_{0}^{2a} 2\sqrt{a}\,\sqrt{x}\,dx\)
\(\displaystyle = 2\sqrt{a}\left[\frac{2}{3}x^{3/2}\right]_{0}^{2a}\)
\(\displaystyle = \frac{4\sqrt{a}}{3}\,(2a)^{3/2}\)
Since \((2a)^{3/2} = 2\sqrt{2}\,a\sqrt{a}\):
\(\displaystyle A = \frac{4\sqrt{a}}{3}\cdot 2\sqrt{2}\,a\sqrt{a} = \frac{8\sqrt{2}\,a^2}{3}\) sq. units
Shortcut: a is a length, so an area must contain \(a^2\). Options (1), (3) and (4) contain \(a^3\), so only Option (2) can be the area.