If the area (in sq. units) of the region bounded by the parabola \(y^{2}=4ax\), a > 0, and the line x = a is 24, then a is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
The region runs from the vertex x = 0 to the line x = a. The upper half of the parabola is \(y = 2\sqrt{a}\,\sqrt{x}\), and the parabola is symmetric about the x-axis, so the area is twice the area above the x-axis:
\( \displaystyle A = 2\int_{0}^{a}2\sqrt{a}\,x^{1/2}\,dx \)
\( \displaystyle = 4\sqrt{a}\left[\frac{2}{3}\,x^{3/2}\right]_{0}^{a} \)
\( \displaystyle = 4\sqrt{a}\times\frac{2}{3}\,a^{3/2} \)
\( \displaystyle = \frac{8a^{2}}{3} \)
Setting this equal to 24:
\( \displaystyle \frac{8a^{2}}{3} = 24 \)
\(a^{2} = 9\)
\(a = 3\) (since a > 0)
Shortcut: The area cut off from a parabola by a chord perpendicular to its axis is two-thirds of the enclosing rectangle. The line x = a meets the parabola at \(y = \pm 2a\), so the rectangle is a by 4a:
\( \displaystyle A = \frac{2}{3}\times 4a^{2} = \frac{8a^{2}}{3} \)