Question

The curve \(y = 4x^2\) and \(y^2 = 2x\) meet at the origin O and at a point P, forming a loop. The straight line OP divides the loop into two parts. What is the ratio of the areas of the two parts of the loop?

• A. 3 : 1
• B. 3 : 2
• C. 2 : 1
• D. 1 : 1

Hint:

Find where the two curves meet, write both curves as x in terms of y, then integrate to compare the area on each side of the chord OP.

Answer 1

The Correct Option is D

Step 1: Find the point P where the two curves meet.
The curves are \(y = 4x^2\) and \(y^2 = 2x\). Substitute \(y = 4x^2\) into the second equation:
\[ (4x^2)^2 = 2x \]
\[ 16x^4 = 2x \]
\[ 16x^4 - 2x = 0 \]
\[ 2x(8x^3 - 1) = 0 \]
This gives \(x = 0\) (the origin O) or \(x^3 = \frac{1}{8}\), so \(x = \frac{1}{2}\).
At \(x = \frac{1}{2}\): \(y = 4\left(\frac{1}{2}\right)^2 = 4 \times \frac{1}{4} = 1\).
So \(P = \left(\frac{1}{2}, 1\right)\).

Step 2: Find the equation of the line OP.
The slope of OP is \(\dfrac{1-0}{\frac{1}{2}-0} = 2\), so the line is \(y = 2x\), or equivalently \(x = \dfrac{y}{2}\).

Step 3: Express both curves as functions of \(y\), since this makes the loop easy to integrate from \(y=0\) to \(y=1\).
From \(y = 4x^2\): \(x = \dfrac{\sqrt{y}}{2}\) (taking the positive root, since the loop lies where \(x \ge 0\)).
From \(y^2 = 2x\): \(x = \dfrac{y^2}{2}\).
For \(0 < y < 1\), we have \(y^2 < y < \sqrt{y}\), so \(\dfrac{y^2}{2} < \dfrac{\sqrt{y}}{2}\). This means the curve \(y=4x^2\) lies to the right and \(y^2=2x\) lies to the left, and together they enclose the loop.

Step 4: Find the total area of the loop.
\[ \text{Loop Area} = \int_0^1 \left(\frac{\sqrt{y}}{2} - \frac{y^2}{2}\right) dy = \frac{1}{2}\left[\int_0^1 \sqrt{y}\,dy - \int_0^1 y^2\,dy\right] \]
\[ \int_0^1 \sqrt{y}\,dy = \left[\frac{2}{3}y^{3/2}\right]_0^1 = \frac{2}{3}, \qquad \int_0^1 y^2\,dy = \left[\frac{y^3}{3}\right]_0^1 = \frac{1}{3} \]
\[ \text{Loop Area} = \frac{1}{2}\left(\frac{2}{3}-\frac{1}{3}\right) = \frac{1}{2}\times\frac{1}{3} = \frac{1}{6} \]

Step 5: Find the area between line OP and the right-hand curve \(y = 4x^2\).
\[ \text{Area}_1 = \int_0^1 \left(\frac{\sqrt{y}}{2} - \frac{y}{2}\right) dy = \frac{1}{2}\left(\int_0^1 \sqrt{y}\,dy - \int_0^1 y\,dy\right) \]
\[ \int_0^1 y\,dy = \left[\frac{y^2}{2}\right]_0^1 = \frac{1}{2} \]
\[ \text{Area}_1 = \frac{1}{2}\left(\frac{2}{3}-\frac{1}{2}\right) = \frac{1}{2}\times\frac{1}{6} = \frac{1}{12} \]

Step 6: Find the area between line OP and the left-hand curve \(y^2 = 2x\).
\[ \text{Area}_2 = \int_0^1 \left(\frac{y}{2} - \frac{y^2}{2}\right) dy = \frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right) = \frac{1}{2}\times\frac{1}{6} = \frac{1}{12} \]

Step 7: Compare the two parts.
Both parts of the loop have area \(\dfrac{1}{12}\), and together they add up to \(\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{6}\), matching the total loop area from Step 4, which confirms the split is exact.
\[ \text{Area}_1 : \text{Area}_2 = \frac{1}{12} : \frac{1}{12} = 1:1 \]
The other options (3:1, 3:2, 2:1) would come from computing only one part of the loop or from an arithmetic slip in one of the integrals.

Final Answer:
The straight line OP divides the loop into two equal parts. \[ \boxed{1:1} \]