The area (in square units) of the region bounded by the curve \( y^{2}=4ax\ (a\gt 0) \) and the lines \( y=2a,\ x=0 \) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{2a^{2}}{3} \)
The line y = 2a meets the parabola where \( 4a^{2}=4ax \), that is at (a, 2a). The region lies between the y-axis (x = 0) and the parabola, from y = 0 to y = 2a.
In terms of y, the parabola is \( x=\frac{y^{2}}{4a} \), so the area is found by integrating with respect to y:
\( \displaystyle \text{Area}=\int_{0}^{2a}\frac{y^{2}}{4a}\,dy \)
\( \displaystyle =\frac{1}{4a}\left[\frac{y^{3}}{3}\right]_{0}^{2a} \)
\( \displaystyle =\frac{1}{4a}\cdot\frac{8a^{3}}{3} \)
\( \displaystyle =\frac{2a^{2}}{3} \)
Shortcut: The arc of the parabola from the vertex (0, 0) to (a, 2a) splits the rectangle with vertices (0, 0), (a, 0), (a, 2a), (0, 2a), of area \( a\times 2a=2a^{2} \), into two parts. The part under the arc (next to the x-axis) is \( \frac{2}{3} \) of the rectangle, so the part between the arc and the y-axis is the remaining \( \frac{1}{3} \):
\( \displaystyle \text{Area}=\frac{1}{3}\times 2a^{2}=\frac{2a^{2}}{3} \)