The region bounded by the parabola \(x^{2} = 2y\) and the line y = 8 is divided into two parts by the line y = 2. The ratio of the area of the smaller part to the area of the larger part is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 : 7
At height y the parabola has \(x = \pm\sqrt{2y}\), so the region between \(x^{2} = 2y\) and a horizontal line y = k has width \(2\sqrt{2y}\) at height y. Integrating in y:
\( \displaystyle \int_{0}^{k}2\sqrt{2}\,\sqrt{y}\,dy = \left[\frac{4\sqrt{2}}{3}y^{3/2}\right]_{0}^{k} = \frac{4\sqrt{2}}{3}k^{3/2} \)
The part below y = 2 (k = 2, \(2^{3/2} = 2\sqrt{2}\)):
\( \displaystyle A_{1} = \frac{4\sqrt{2}}{3}\times2\sqrt{2} = \frac{16}{3} \)
The whole region (k = 8, \(8^{3/2} = 16\sqrt{2}\)):
\( \displaystyle \frac{4\sqrt{2}}{3}\times16\sqrt{2} = \frac{128}{3} \)
The part between y = 2 and y = 8:
\( \displaystyle A_{2} = \frac{128}{3}-\frac{16}{3} = \frac{112}{3} \)
\( \displaystyle A_{1} : A_{2} = \frac{16}{3} : \frac{112}{3} = 1 : 7 \)
Why the other options are incorrect
Option (1) 1 : 8 — Incorrect. This compares the smaller part with the whole region, \(\frac{16}{3} : \frac{128}{3}\), instead of with the larger part.
Option (2) 1 : 3 — Incorrect. This is the ratio of the heights 2 and 8 − 2; the area grows as \(k^{3/2}\), not as k.
Option (4) 1 : 4 — Incorrect. This is the ratio of the heights 2 and 8; the parts have areas \(\frac{16}{3}\) and \(\frac{112}{3}\).