Equating Enclosed Areas Between Parabolas and a Line to Determine a Parameter

It the area enclosed by the parabolas P_1: 2 y=5 x^2 and P_2: x^2-y+6=0 is equal to the area enclosed by P_1 and y=α x, α>0, then α^3 is equal to

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Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Why can we integrate from 0 to 2 and multiply by 2?

Both parabolas are symmetric about the y-axis because they are even functions of x. Therefore, the region from x = -2 to 0 is the mirror image of the region from x = 0 to 2.

Answer

Accepted answer: 600

The value of $α^{3}$ is equal to 600.

Why each option works or fails

Step-by-step solution

Given: Parabolas $P_{1} : y = \frac{5}{2} x^{2}$ and $P_{2} : y = x^{2} + 6$ , line $y = α x$ with $α > 0$ , and $Area (P_{1}, P_{2}) = Area (P_{1}, y = α x)$ .
Goal: Find the value of $α^{3}$ .
Approach: First, find the intersection points of $P_{1}$ and $P_{2}$ to calculate the enclosed area $A_{1}$ using integration. Then, determine the intersection of $P_{1}$ with the line $y = α x$ to find $A_{2}$ as a function of $α$ . Finally, equate $A_{1} = A_{2}$ to solve for $α^{3}$ .
Execute: Find intersection of $P_{1}$ and $P_{2}$ : $\frac{5}{2} x^{2} = x^{2} + 6 ⟹ \frac{3}{2} x^{2} = 6 ⟹ x^{2} = 4 ⟹ x = \pm 2$ . Compute area $A_{1} = \int_{- 2}^{2} (x^{2} + 6 - \frac{5}{2} x^{2}) d x = 2 \int_{0}^{2} (6 - \frac{3}{2} x^{2}) d x = 2 {[6 x - \frac{x^{3}}{2}]}_{0}^{2} = 2 (12 - 4) = 16$ .
Execute: Find intersection of $y = \frac{5}{2} x^{2}$ and $y = α x$ : $\frac{5}{2} x^{2} = α x ⟹ x (5 x - 2 α) = 0$ , so $x = 0$ and $x = \frac{2 α}{5}$ . Compute enclosed area $A_{2} = \int_{0}^{2 α / 5} (α x - \frac{5}{2} x^{2}) d x = {[\frac{α x^{2}}{2} - \frac{5 x^{3}}{6}]}_{0}^{2 α / 5} = \frac{α}{2} (\frac{4 α^{2}}{25}) - \frac{5}{6} (\frac{8 α^{3}}{125}) = \frac{2 α^{3}}{25} - \frac{4 α^{3}}{75} = \frac{2 α^{3}}{75}$ .
Execute: Equate $A_{1} = A_{2}$ : $\frac{2}{75} α^{3} = 16 ⟹ α^{3} = 16 \times \frac{75}{2} = 8 \times 75 = 600$ .
Verify: Check formula for area between $y^{2} = 4 a x$ and $y = m x$ or $x^{2} = 4 b y$ and $y = m x$ : for $x^{2} = \frac{2}{5} y$ (so $4 b = 2 / 5 ⟹ b = 1 / 10$ ) and $y = m x$ , the area is $\frac{8}{3} b^{2} m^{3} = \frac{8}{3} \cdot \frac{1}{100} \cdot α^{3} = \frac{2}{75} α^{3}$ . Setting $\frac{2}{75} α^{3} = 16$ gives $α^{3} = 600$ . Consistent and verified.

Equating Enclosed Areas Between Parabolas and a Line to Determine a Parameter

What feels right?

The value of $α^{3}$ is equal to 600.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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