Let P1: y = 4x2 and P2: y = x2 + 27 be two parabolas. If the area of the bounded region enclosed between P1 and P2 is six times the area of the bounded region enclosed between the line y = alpha x, alpha > 0 and P1, then alpha is equal to:

Area enclosed by y = f(x) and y = g(x) from x=a to x=b: Area between x2 = 4ay and line x = my is (8a2/3m3). Find Points of Intersection for P1 and P2: Area bounded between P1 and P2: Area between P1 and y = alpha x is given as 108 / 6 = 18 sq. units. For x2 = 4ay and x = my, area is (8a2/3m3). Here, the parabola is x2 = (y/4) and line is x = (y/alpha). Thus, 4a = (1/4) a = (1/16), and m = (1/alpha). Applying the formula: alpha = 12 Using the standard result (8a2/3m3) for the area bounded between x2=4ay and x=my avoids tedious integration for straight lines intersecting parabolas at the origin. Class 12 Maths: Application of Integrals

Let P1: y = 4x2 and P2: y = x2 + 27 be two parabolas. If the area of the bounded region enclosed between P1 and P2 is six times the area of the bounded region enclosed between the line y = alpha x, al

Area enclosed by y = f(x) and y = g(x) from x=a to x=b: Area between x2 = 4ay and line x = my is (8a2/3m3). Find Points of Intersection for P1 and P2: Area bounded between P1 and P2: Area between P1 and y = alpha x is given as 108 / 6 = 18 sq. units. For x2 = 4ay and x = my, area is (8a2/3m3). Here, the parabola is x2 = (y/4) and line is x = (y/alpha). Thus, 4a = (1/4) a = (1/16), and m = (1/alpha). Applying the formula: alpha = 12 Using the standard result (8a2/3m3) for the area bounded between

Let P1: y = 4x2 and P2: y = x2 + 27 be two parabolas. If the area of the bounded region enclosed between P1 and P2 is six times the area of the bounded region enclosed between the line y = alpha x, alpha > 0 and P1, then alpha is equal to:

Area enclosed by y = f(x) and y = g(x) from x=a to x=b: Area between x2 = 4ay and line x = my is (8a2/3m3). Find Points of Intersection for P1 and P2: Area bounded between P1 and P2: Area between P1 and y = alpha x is given as 108 / 6 = 18 sq. units. For x2 = 4ay and x = my, area is (8a2/3m3). Here, the parabola is x2 = (y/4) and line is x = (y/alpha). Thus, 4a = (1/4) a = (1/16), and m = (1/alpha). Applying the formula: alpha = 12 Using the standard result (8a2/3m3) for the area bounded between x2=4ay and x=my avoids tedious integration for straight lines intersecting parabolas at the origin. Class 12 Maths: Application of Integrals

Let P1: y = 4x2 and P2: y = x2 + 27 be two parabolas. If the area of the bounded region enclosed between P1 and P2 is six times the area of the bounded region enclosed between the line y = alpha x, al

Area enclosed by y = f(x) and y = g(x) from x=a to x=b: Area between x2 = 4ay and line x = my is (8a2/3m3). Find Points of Intersection for P1 and P2: Area bounded between P1 and P2: Area between P1 and y = alpha x is given as 108 / 6 = 18 sq. units. For x2 = 4ay and x = my, area is (8a2/3m3). Here, the parabola is x2 = (y/4) and line is x = (y/alpha). Thus, 4a = (1/4) a = (1/16), and m = (1/alpha). Applying the formula: alpha = 12 Using the standard result (8a2/3m3) for the area bounded between

Let P1: y = 4x2 and P2: y = x2 + 27 be two parabolas. If the area of the bounded region enclosed between P1 and P2 is…

Year	2026	2025	2024	Total
Questions	9	11	7	27

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Area enclosed by $y = f(x)$ $y = f(x)$ and $y = g(x)$ $y = g(x)$ from $x=a$ $x=a$ to $x=b$ $x=b$:

A = \intₐ^b |f(x) - g(x)| dx

Area between $x² = 4ay$ $x^2 = 4ay$ and line $x = my$ $x = my$ is $(8a²)/(3m³)$ $\frac{8a^2}{3m^3}$.

Core Logic

Find Points of Intersection for $P₁$ $P_1$ and $P₂$ $P_2$:

4x² = x² + 27 \Rightarrow 3x² = 27 \Rightarrow x = \pm 3

Area bounded between $P₁$ $P_1$ and $P₂$ $P_2$:

A₁ = \int₋₃³ ((x² + 27) - 4x²) dx = 2 \int₀³ (27 - 3x²) dx

A₁ = 2 [27x - x³]₀³ = 2(81 - 27) = 108 sq. units

Execution

Area between $P₁$ $P_1$ and $y = α x$ $y = \alpha x$ is given as $108 / 6 = 18$ $108 / 6 = 18$ sq. units. For $x² = 4ay$ $x^2 = 4ay$ and $x = my$ $x = my$, area is $(8a²)/(3m³)$ $\frac{8a^2}{3m^3}$. Here, the parabola is $x² = (y)/(4)$ $x^2 = \frac{y}{4}$ and line is $x = (y)/(α)$ $x = \frac{y}{\alpha}$. Thus, $4a = (1)/(4) \Rightarrow a = (1)/(16)$ , and $m = (1)/(α)$ $m = \frac{1}{\alpha}$.

Applying the formula:

(8(1/16)²)/(3(1/α)³) = 18

(8 / 256)/(3 / α³) = 18 \Rightarrow (α³)/(96) = 18

α³ = 18 \times 96 = 1728

$α = 12$ $\alpha = 12$

Pattern Recognition

Using the standard result $(8a²)/(3m³)$ $\frac{8a^2}{3m^3}$ for the area bounded between $x²=4ay$ $x^2=4ay$ and $x=my$ $x=my$ avoids tedious integration for straight lines intersecting parabolas at the origin.

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

Solution & Explanation

Related Formula

Core Logic

Execution

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

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Solution

Core Logic

Step 1: Case I (y > 0)

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