The area of the region bounded by the parabola \(y^2 = 8x\) and its latus rectum in the first quadrant, is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{16}{3}\) sq. units
Comparing \(y^2 = 8x\) with \(y^2 = 4ax\) gives \(a = 2\). The latus rectum is the line \(x = a = 2\) through the focus.
In the first quadrant \(y = \sqrt{8x} = 2\sqrt{2}\,\sqrt{x}\), and the region runs from \(x = 0\) to \(x = 2\):
\( \displaystyle A = \int_{0}^{2} 2\sqrt{2}\,\sqrt{x}\,dx = 2\sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_{0}^{2} \)
\( \displaystyle = 2\sqrt{2} \times \frac{2}{3} \times 2\sqrt{2} \)
\( \displaystyle = \frac{16}{3} \text{ sq. units} \)
Tip: The question asks only for the part in the first quadrant. The whole region between the parabola and its latus rectum, on both sides of the x-axis, is twice this, \(\frac{32}{3}\), which is Option (2).