The area (in square units) of the region bounded by the parabola y2 = 4(x - 2) and the line y = 2x

Year	2026	2025	2024	Total
Questions	9	11	7	27

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Area = \inty₁y₂ (xR - xL) dy

Core Logic

Area under Curves diagram for Q8 - JEE Main 2024 Morning

To simplify calculations, shift the origin. Let $X = x - 2$ $X = x - 2$. The equations become: Parabola: $y² = 4X \Rightarrow X = (y²)/(4)$ Line: $y = 2(X + 2) - 8 \Rightarrow y = 2X - 4 \Rightarrow X = (y + 4)/(2)$

Step 1: Finding points of intersection

Set the $X$ $X$ values equal to find intersection points in terms of $y$ $y$:

(y²)/(4) = (y + 4)/(2)

$y² = 2y + 8$ $y^2 = 2y + 8$

y² - 2y - 8 = 0

(y - 4)(y + 2) = 0

The intersection points are at $y = -2$ $y = -2$ and $y = 4$ $y = 4$.

Step 2: Area Integration

Integrate with respect to $y$ $y$ from $-2$ $-2$ to $4$ $4$:

A = \int₋₂⁴ ( xR - xL ) dy

A = \int₋₂⁴ ( (y + 4)/(2) - (y²)/(4) ) dy

A = [ (y²)/(4) + 2y - (y³)/(12) ]₋₂⁴

Upper limit $(y=4)$ $(y=4)$: $(16)/(4) + 8 - (64)/(12) = 4 + 8 - (16)/(3) = 12 - (16)/(3) = (20)/(3)$ Lower limit $(y=-2)$ $(y=-2)$: $(4)/(4) - 4 - (-8)/(12) = 1 - 4 + (2)/(3) = -3 + (2)/(3) = -(7)/(3)$

A = (20)/(3) - (-(7)/(3)) = (27)/(3) = 9

The solution simplifies it directly to $9$ $9$ square units.

Pattern Recognition

For a horizontal parabola interacting with a line, integrating along the y-axis is always cleaner than splitting it into multiple integrals along the x-axis.

Chapter Mix

Class 12 Maths: Application of Integrals

More Application of Integrals Previous-Year Questions — Page 6

Q11 jee_main_2024_31_jan_evening Area Under Curves

The area of the region enclosed by the parabola

y = 4x - x²

and

3y = (x - 4)²

is equal to

A. $(32)/(9)$ $\frac{32}{9}$
B. $4$ $4$
C. $6$ $6$
D. $(14)/(3)$ $\frac{14}{3}$

Solution

Related Formula

Area = \intₐb (yupper - ylower) dx

Core Logic

Area Under Curves diagram for Q11 - JEE Main 2024 Evening

Find intersection points of $y = 4x - x²$ $y = 4x - x^2$ and $3y = (x - 4)²$ $3y = (x - 4)^2$:

3(4x - x²) = x² - 8x + 16

12x - 3x² = x² - 8x + 16

4x² - 20x + 16 = 0 x² - 5x + 4 = 0

Roots are $x = 1, 4$ $x = 1, 4$.

Area integral:

Area = \int₁⁴ [ (4x - x²) - ((x - 4)²)/(3) ] dx

= [ (4x²)/(2) - (x³)/(3) - ((x - 4)³)/(9) ]₁⁴

= [ 2(16) - (64)/(3) - 0 ] - [ 2(1) - (1)/(3) - ((-3)³)/(9) ]

= ( 32 - (64)/(3) ) - ( 2 - (1)/(3) + 3 )

= (32)/(3) - ( 5 - (1)/(3) ) = (32)/(3) - (14)/(3) = (18)/(3) = 6

Chapter Mix

Class 12 Maths: Applications of the Integrals

Q5 jee_main_2024_31_jan_morning Area bounded by Parabolas and Inequalities

The area of the region

(x,y): y² \leq 4x, x < 4, (xy(x - 1)(x - 2))/((x - 3)(x - 4)) > 0, x \neq 3

A. $(16)/(3)$ $$\frac{16}{3}$$
B. $(64)/(3)$ $$\frac{64}{3}$$
C. $(8)/(3)$ $$\frac{8}{3}$$
D. $(32)/(3)$ $$\frac{32}{3}$$

Solution

Core Logic

Given $y² \leq 4x$ $y^2 \le 4x$ and $x < 4$ $x < 4$. Analyze the inequality $(xy(x-1)(x-2))/((x-3)(x-4)) > 0$ $\frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0$ considering $y > 0$ $y > 0$ and $y < 0$ $y < 0$ separately.

Area bounded by Parabolas and Inequalities diagram for Q5 - JEE Main 2024 Morning

Step 1: Case I (y > 0)

If $y > 0$ $y > 0$, the inequality reduces to $(x(x-1)(x-2))/((x-3)(x-4)) > 0$ $\frac{x(x-1)(x-2)}{(x-3)(x-4)} > 0$. Using wavy curve method and given $x in (0, 4)$ $x \in (0, 4)$: $x in (0, 1) (2, 3)$ $x \in (0, 1) \cup (2, 3)$.

Step 2: Case II (y < 0)

If $y < 0$ $y < 0$, the inequality reduces to $(x(x-1)(x-2))/((x-3)(x-4)) < 0$ $\frac{x(x-1)(x-2)}{(x-3)(x-4)} < 0$. Using wavy curve method and given $x in (0, 4)$ $x \in (0, 4)$: $x in (1, 2) (3, 4)$ $x \in (1, 2) \cup (3, 4)$.

Step 3: Area Computation

Because the regions map perfectly without overlap in opposite quadrants relative to the x-axis, they form complete parabolic strips when combined: Area $= 2 \int₀⁴ \sqrt(x) dx = 2 \cdot (2)/(3)[x3/2]₀⁴ = (4)/(3) \cdot 8 = (32)/(3)$ .

Chapter Mix

Class 12 Maths: Area Under Curves Class 11 Maths: Linear Inequalities

The area (in square units) of the region bounded by the parabola $y² = 4(x - 2)$ $y^2 = 4(x - 2)$ and the line $y = 2x - 8$ $y = 2x - 8$

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding points of intersection

Step 2: Area Integration

Pattern Recognition

Chapter Mix

Reference Study Guides

More Application of Integrals Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Case I (y > 0)

Step 2: Case II (y < 0)

Step 3: Area Computation

Chapter Mix

More Application of Integrals Questions — jee_main_2024_30_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

The area (in square units) of the region bounded by the parabola y² = 4(x - 2)$y^2 = 4(x - 2)$ and the line y = 2x - 8$y = 2x - 8$

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding points of intersection

Step 2: Area Integration

Pattern Recognition

Chapter Mix

Reference Study Guides

More Application of Integrals Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Case I (y > 0)

Step 2: Case II (y < 0)

Step 3: Area Computation

Chapter Mix

More Application of Integrals Questions — jee_main_2024_30_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

The area (in square units) of the region bounded by the parabola $y² = 4(x - 2)$ $y^2 = 4(x - 2)$ and the line $y = 2x - 8$ $y = 2x - 8$