A line passing through the point mathrmA(-2, 0)$\mathrm{A}(-2, 0)$, touches the parabola mathrmP: mathrmy^2 = mathrmx - 2$\mathrm{P}: \mathrm{y}^2 = \mathrm{x} - 2$ at the point mathrmB$\mathrm{B}$ in the first quadrant. The area, of the region bounded by the line mathrmAB$\mathrm{AB}$, parabola mathrmP$\mathrm{P}$ and the mathbfx$\mathbf{x}$-axis, is :-
A.frac73$\frac{7}{3}$
B.2$2$
C.frac83$\frac{8}{3}$
D.3$3$
Solution & Explanation
### Core Logic
Let the equation of the tangent line passing through A(-2,0)$A(-2,0)$ be:
y = m(x + 2) implies x = fracym - 2$y = m(x + 2) \implies x = \frac{y}{m} - 2$
The equation of the parabola is y^2 = x - 2 implies x = y^2 + 2$y^2 = x - 2 \implies x = y^2 + 2$.
Substituting x$x$ from the line into the parabola:
y^2 + 2 = fracym - 2 implies y^2 - fracym + 4 = 0$y^2 + 2 = \frac{y}{m} - 2 \implies y^2 - \frac{y}{m} + 4 = 0$
For the line to be a tangent, the discriminant of this quadratic equation must be zero (D = 0$D = 0$):
left(-frac1mright)^2 - 4(1)(4) = 0 implies frac1m^2 = 16 implies m = pm frac14$\left(-\frac{1}{m}\right)^2 - 4(1)(4) = 0 \implies \frac{1}{m^2} = 16 \implies m = \pm \frac{1}{4}$
Since point B$B$ is in the first quadrant, the slope must be positive, so m = frac14$m = \frac{1}{4}$.
The line equation is y = frac14(x + 2) implies x = 4y - 2$y = \frac{1}{4}(x + 2) \implies x = 4y - 2$.
The point of tangency B$B$ is found at y = frac12m = 2$y = \frac{1}{2m} = 2$, which gives x = 6$x = 6$, so B = (6,2)$B = (6,2)$.
### Step 1: Setting up the Area Integral
Integrating with respect to y$y$ avoids splitting the region into two parts along the x-axis:
textArea = int_0^2 left(x_textparabola - x_textlineright) dy$\text{Area} = \int_{0}^{2} \left(x_{\text{parabola}} - x_{\text{line}}\right) dy$textArea = int_0^2 left((y^2 + 2) - (4y - 2)right) dy = int_0^2 left(y^2 - 4y + 4right) dy$\text{Area} = \int_{0}^{2} \left((y^2 + 2) - (4y - 2)\right) dy = \int_{0}^{2} \left(y^2 - 4y + 4\right) dy$Area under curves diagram for Q59 - JEE Main 2025 Evening
### Step 2: Evaluating the Integral
Integrating term by term:
textArea = left[ fracy^33 - 2y^2 + 4y right]_0^2$\text{Area} = \left[ \frac{y^3}{3} - 2y^2 + 4y \right]_{0}^{2}$textArea = left( frac83 - 2(4) + 4(2) right) - 0 = frac83 - 8 + 8 = frac83$\text{Area} = \left( \frac{8}{3} - 2(4) + 4(2) \right) - 0 = \frac{8}{3} - 8 + 8 = \frac{8}{3}$
### Pattern Recognition
Integrating with respect to y$y$ (horizontal strips) when dealing with horizontal parabolas or lines crossing the x-axis eliminates the need to break your area computation into multiple piecewise integrals.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Area Under Curves
Class 11 Mathematics: Conic Sections
Keywords:#area bounded by line and parabola x-axis#JEE Main 2025 Evening Q59#Area Under Curves JEE Main 2025#Area Bounded by Parabola and Tangent JEE Main 2025
More Area Under Curves Previous-Year Questions
Q702025Area Bounded by Curves and Absolute Value Functions
The area (in sq. units) of the region
\(x, y): 0 le y le 2|x| + 1, 0 le y le x^2 + 1, |x| le 3\$\{(x, y): 0 \le y \le 2|x| + 1, 0 \le y \le x^2 + 1, |x| \le 3\}$
is
(1) frac803$\frac{80}{3}$
(2) frac643$\frac{64}{3}$
(3) frac173$\frac{17}{3}$
(4) frac323$\frac{32}{3}$
A.frac803$\frac{80}{3}$
B.frac643$\frac{64}{3}$
C.frac173$\frac{17}{3}$
D.frac323$\frac{32}{3}$
Solution
### Related Formula
Area under a curve using definite integration bounded by curves:
textArea = 2 int_a^b f(x) dx$\text{Area} = 2 \int_a^b f(x) dx$
### Core Logic
Since both bounding graphs are symmetric about the y-axis, we can integrate over the positive domain (x ge 0$x \ge 0$) and double the result: Area Bounded by Curves and Absolute Value Functions diagram for Q70 - JEE Main 2025 Morning
Intersecting points: x^2 + 1 = 2x + 1 implies x = 2$x^2 + 1 = 2x + 1 \implies x = 2$.
### Step 1: Setting Up the Bound Segments
The region splits into two integral segments based on which curve sits lower:
Segment 1: From 0$0$ to 2$2$, bounded by the parabola y = x^2 + 1$y = x^2 + 1$.
Segment 2: From 2$2$ to 3$3$, bounded by the straight line y = 2x + 1$y = 2x + 1$.
### Step 2: Performing Integration
$textArea = 2 left[ int_0^2 (x^2 + 1) dx + int_2^3 (2x + 1) dx right]$$\text{Area} = 2 \left[ \int_0^2 (x^2 + 1) dx + \int_2^3 (2x + 1) dx \right]$$
$= 2 left[ left( frac83 + 2 right) + (9 + 3 - 4 - 2) right] = 2 left[ frac143 + 6 right] = frac643$= 2 \left[ \left( \frac{8}{3} + 2 \right) + (9 + 3 - 4 - 2) \right] = 2 \left[ \frac{14}{3} + 6 \right] = \frac{64}{3}$$
### Pattern Recognition
Exploiting structural symmetry drops the integration limits, avoiding messy sign evaluations with absolute terms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Area Under Curves
Q712025Area Bounded by Multiple Curves
If the area of the region \(x,y) : |x - 5| le y le 4sqrtx\$\{(x,y) : |x - 5| \le y \le 4\sqrt{x}\}$ is A, then 3A is equal to
Numerical Answer.Answer: 368 to 368
Solution
### Related Formula
Area tracking equation via horizontal slices or split verticals:
textArea = int_x_1^x_2 (y_textupper - y_textlower)\,mathrmdx$\text{Area} = \int_{x_1}^{x_2} (y_{\text{upper}} - y_{\text{lower}})\,\mathrm{d}x$
### Core Logic
Find the intersections of y = |x - 5|$y = |x - 5|$ and y = 4sqrtx$y = 4\sqrt{x}$:
Branch 1: 5 - x = 4sqrtx implies x + 4sqrtx - 5 = 0 implies (sqrtx + 5)(sqrtx - 1) = 0 implies x = 1, y = 4$5 - x = 4\sqrt{x} \implies x + 4\sqrt{x} - 5 = 0 \implies (\sqrt{x} + 5)(\sqrt{x} - 1) = 0 \implies x = 1, y = 4$.
Branch 2: x - 5 = 4sqrtx implies x - 4sqrtx - 5 = 0 implies (sqrtx - 5)(sqrtx + 1) = 0 implies x = 25, y = 20$x - 5 = 4\sqrt{x} \implies x - 4\sqrt{x} - 5 = 0 \implies (\sqrt{x} - 5)(\sqrt{x} + 1) = 0 \implies x = 25, y = 20$.
Area Bounded by Multiple Curves diagram for Q71 - JEE Main 2025 Morning
### Step 1: Set Up Area Definite Integral
Split integration intervals around vertex x = 5$x = 5$ or compute via simple boundary differences:
A = int_1^25 4sqrtx\,mathrmdx - textArea of Left Triangle - textArea of Right Triangle$A = \int_{1}^{25} 4\sqrt{x}\,\mathrm{d}x - \text{Area of Left Triangle} - \text{Area of Right Triangle}$textArea of Left Triangle = frac12 times (5 - 1) times 4 = 8$\text{Area of Left Triangle} = \frac{1}{2} \times (5 - 1) \times 4 = 8$textArea of Right Triangle = frac12 times (25 - 5) times 20 = 200$\text{Area of Right Triangle} = \frac{1}{2} \times (25 - 5) \times 20 = 200$
### Step 2: Complete Computations
int_1^25 4sqrtx\,mathrmdx = left[ frac83x^3/2 right]_1^25 = frac83(125 - 1) = frac8 times 1243 = frac9923$\int_{1}^{25} 4\sqrt{x}\,\mathrm{d}x = \left[ \frac{8}{3}x^{3/2} \right]_{1}^{25} = \frac{8}{3}(125 - 1) = \frac{8 \times 124}{3} = \frac{992}{3}$A = frac9923 - 208 = frac992 - 6243 = frac3683$A = \frac{992}{3} - 208 = \frac{992 - 624}{3} = \frac{368}{3}$3A = 368$3A = 368$
### Pattern Recognition
Subtracting standard geometric triangles beneath linear configurations from total absolute root curves saves significant time compared to managing multiple separate analytical integral pieces.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Area Under Curves
Q592025Area Bounded by Algebraic and Rational Curves
The area of the region bounded by the curves x(1+y^2)=1$x(1+y^{2})=1$ and y^2=2x$y^{2}=2x$ is :
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