JEE MainMathematics

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Area Enclosed Between a Parabola and a Line

Q: Does this formula work for any line and parabola?

Yes, for any parabola x = Ay^2 + By + C and any line intersecting it at two points with y-coordinates y_1 and y_2, the enclosed area is always (|A|)/6|y_2 - y_1|^3.

The area enclosed by the curves y^2 + 4x = 4 and y - 2x = 2 is:

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A9B22/3C23/3D25/3

Step-by-step solutionView

Correct answer

The area enclosed between the parabola $x = 1 - \frac{y^{2}}{4}$ and the line $x = \frac{y - 2}{2}$ is $9$ .

Option analysis

Why each option works or fails

A · 9

This option correctly evaluates the enclosed area. Integrating with respect to $y$ from $y = - 4$ to $y = 2$ gives $\int_{- 4}^{2} (2 - \frac{y}{2} - \frac{y^{2}}{4}) d y = 9$ .

B · 22/3

Evaluating the integral with an arithmetic sign error in the constant term or using asymmetric integration limits leads to $\frac{22}{3}$ . Ensure the bounds are found by equating $4 - y^{2} = 2 y - 4$ , giving $y^{2} + 2 y - 8 = 0$ , so $y = - 4$ and $y = 2$ .

C · 23/3

Making a slip when evaluating the cubic term ${[- \frac{y^{3}}{12}]}_{- 4}^{2}$ gives a net total of $\frac{23}{3}$ . Check that ${[- \frac{y^{3}}{12}]}_{- 4}^{2} = - \frac{8}{12} - (- \frac{- 64}{12}) = - \frac{8}{12} - \frac{64}{12} = - 6$ .

D · 25/3

Adding an extra $\frac{2}{3}$ or making a sign slip when substituting the lower limit $y = - 4$ yields $\frac{25}{3}$ . Carefully compute each term: $[2 y]_{- 4}^{2} = 12$ , ${[- \frac{y^{2}}{4}]}_{- 4}^{2} = - 3$ , and ${[- \frac{y^{3}}{12}]}_{- 4}^{2} = - 6$ , summing to $12 - 3 - 6 = 9$ .

Reviewed route

Solution

StepWorking

01Given

The boundary curves are the parabola $y^{2} + 4 x = 4$ and the straight line $y - 2 x = 2$ .

02Goal

Find the area of the region enclosed between the parabola and the line.

03Approach

Express $x$ as a function of $y$ for both curves, find their intersection points in terms of $y$ , and integrate $(x_{right} - x_{left}) d y$ from the lower to the upper $y$ -limit.

04Execute

From the line, $2 x = y - 2 ⟹ x = \frac{y - 2}{2}$ . Substitute into the parabola: $y^{2} + 4 (\frac{y - 2}{2}) = 4 ⟹ y^{2} + 2 y - 4 = 4 ⟹ y^{2} + 2 y - 8 = 0$ . Factoring gives $(y + 4) (y - 2) = 0$ , so $y = - 4$ and $y = 2$ .

05Execute

Set up the area integral with $x_{right} = \frac{4 - y^{2}}{4}$ and $x_{left} = \frac{y - 2}{2}$ : $A = \int_{- 4}^{2} (\frac{4 - y^{2}}{4} - \frac{y - 2}{2}) d y = \int_{- 4}^{2} (2 - \frac{y}{2} - \frac{y^{2}}{4}) d y$ . Evaluating the antiderivative: ${[2 y - \frac{y^{2}}{4} - \frac{y^{3}}{12}]}_{- 4}^{2} = (4 - 1 - \frac{8}{12}) - (- 8 - 4 - \frac{- 64}{12}) = \frac{7}{3} - (- 12 + \frac{16}{3}) = \frac{7}{3} - (- \frac{20}{3}) = 9$ .

✓Verify

Using Archimedes' formula for a parabolic segment cut by a chord: $Area = \frac{∣ a ∣}{6} (y_{2} - y_{1})^{3}$ for a parabola $x = - \frac{1}{4} y^{2} + 1$ . Here $∣ a ∣ = \frac{1}{4}$ , $y_{2} - y_{1} = 2 - (- 4) = 6$ . So $Area = \frac{1 / 4}{6} \times 6^{3} = \frac{1}{24} \times 216 = 9$ . The result is verified.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why integrate with respect to y instead of x?

Integrating with respect to x requires splitting the region into two parts because the upper and lower boundary curves change at x = 0. Integrating with respect to y requires just a single integral.

Does this formula work for any line and parabola?

Yes, for any parabola $x = A y^{2} + B y + C$ and any line intersecting it at two points with y-coordinates $y_{1}$ and $y_{2}$ , the enclosed area is always $\frac{∣ A ∣}{6} ∣ y_{2} - y_{1} ∣^{3}$ .

Answer

The area enclosed between the parabola $x = 1 - \frac{y^{2}}{4}$ and the line $x = \frac{y - 2}{2}$ is $9$ .

Why each option works or fails

A · correct: 9 - This option correctly evaluates the enclosed area. Integrating with respect to $y$ from $y = - 4$ to $y = 2$ gives $\int_{- 4}^{2} (2 - \frac{y}{2} - \frac{y^{2}}{4}) d y = 9$ .
B: 22/3 - Evaluating the integral with an arithmetic sign error in the constant term or using asymmetric integration limits leads to $\frac{22}{3}$ . Ensure the bounds are found by equating $4 - y^{2} = 2 y - 4$ , giving $y^{2} + 2 y - 8 = 0$ , so $y = - 4$ and $y = 2$ .
C: 23/3 - Making a slip when evaluating the cubic term ${[- \frac{y^{3}}{12}]}_{- 4}^{2}$ gives a net total of $\frac{23}{3}$ . Check that ${[- \frac{y^{3}}{12}]}_{- 4}^{2} = - \frac{8}{12} - (- \frac{- 64}{12}) = - \frac{8}{12} - \frac{64}{12} = - 6$ .
D: 25/3 - Adding an extra $\frac{2}{3}$ or making a sign slip when substituting the lower limit $y = - 4$ yields $\frac{25}{3}$ . Carefully compute each term: $[2 y]_{- 4}^{2} = 12$ , ${[- \frac{y^{2}}{4}]}_{- 4}^{2} = - 3$ , and ${[- \frac{y^{3}}{12}]}_{- 4}^{2} = - 6$ , summing to $12 - 3 - 6 = 9$ .

Step-by-step solution

Given: The boundary curves are the parabola $y^{2} + 4 x = 4$ and the straight line $y - 2 x = 2$ .
Goal: Find the area of the region enclosed between the parabola and the line.
Approach: Express $x$ as a function of $y$ for both curves, find their intersection points in terms of $y$ , and integrate $(x_{right} - x_{left}) d y$ from the lower to the upper $y$ -limit.
Execute: From the line, $2 x = y - 2 ⟹ x = \frac{y - 2}{2}$ . Substitute into the parabola: $y^{2} + 4 (\frac{y - 2}{2}) = 4 ⟹ y^{2} + 2 y - 4 = 4 ⟹ y^{2} + 2 y - 8 = 0$ . Factoring gives $(y + 4) (y - 2) = 0$ , so $y = - 4$ and $y = 2$ .
Execute: Set up the area integral with $x_{right} = \frac{4 - y^{2}}{4}$ and $x_{left} = \frac{y - 2}{2}$ : $A = \int_{- 4}^{2} (\frac{4 - y^{2}}{4} - \frac{y - 2}{2}) d y = \int_{- 4}^{2} (2 - \frac{y}{2} - \frac{y^{2}}{4}) d y$ . Evaluating the antiderivative: ${[2 y - \frac{y^{2}}{4} - \frac{y^{3}}{12}]}_{- 4}^{2} = (4 - 1 - \frac{8}{12}) - (- 8 - 4 - \frac{- 64}{12}) = \frac{7}{3} - (- 12 + \frac{16}{3}) = \frac{7}{3} - (- \frac{20}{3}) = 9$ .
Verify: Using Archimedes' formula for a parabolic segment cut by a chord: $Area = \frac{∣ a ∣}{6} (y_{2} - y_{1})^{3}$ for a parabola $x = - \frac{1}{4} y^{2} + 1$ . Here $∣ a ∣ = \frac{1}{4}$ , $y_{2} - y_{1} = 2 - (- 4) = 6$ . So $Area = \frac{1 / 4}{6} \times 6^{3} = \frac{1}{24} \times 216 = 9$ . The result is verified.

Shortcut:

When to use it: Whenever finding the area enclosed between any parabola and any secant line.

Given: Parabola $x = 1 - \frac{1}{4} y^{2}$ (leading coefficient in $y$ is $A = - 1 / 4$ ) intersected by the chord from $y = - 4$ to $y = 2$ .

Approach: The area between a parabola $x = A y^{2} + B y + C$ and a secant line intersecting it at $y_{1}, y_{2}$ is given by $\frac{∣ A ∣}{6} (y_{2} - y_{1})^{3}$ .

Execute: Here $∣ A ∣ = \frac{1}{4}$ , $y_{1} = - 4$ , $y_{2} = 2$ . The difference is $y_{2} - y_{1} = 6$ . Therefore, $Area = \frac{1}{6} \times \frac{1}{4} \times 6^{3} = \frac{216}{24} = 9$ .

Verify: Since $\frac{216}{24} = 9$ , the computation matches the standard integration method.

Area Enclosed Between a Parabola and a Line

What feels right?

The area enclosed between the parabola $x = 1 - \frac{y^{2}}{4}$ and the line $x = \frac{y - 2}{2}$ is $9$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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