If the area (in sq. units) of the region bounded by the parabola \(y = ax^{2}\), where a > 0, and the line y = 9 is 12, then a is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9
The parabola meets the line where
\(ax^{2} = 9\)
\( \displaystyle x = \pm\frac{3}{\sqrt{a}} \)
Between these points the line is above the parabola, and the region is symmetric about the y-axis:
\( \displaystyle A = 2\int_{0}^{3/\sqrt{a}}(9-ax^{2})\,dx \)
\( \displaystyle = 2\left[9x-\frac{ax^{3}}{3}\right]_{0}^{3/\sqrt{a}} \)
\( \displaystyle = 2\left(\frac{27}{\sqrt{a}}-\frac{9}{\sqrt{a}}\right) = \frac{36}{\sqrt{a}} \)
Setting this equal to 12:
\( \displaystyle \frac{36}{\sqrt{a}} = 12 \)
\(\sqrt{a} = 3\)
\(a = 9\)
Shortcut: A region cut off from a parabola by a line perpendicular to its axis has area two-thirds of the enclosing rectangle. Here the rectangle is \(\displaystyle\frac{6}{\sqrt{a}}\) wide and 9 high, so \(\displaystyle A = \frac{2}{3}\times\frac{54}{\sqrt{a}} = \frac{36}{\sqrt{a}}\).