The area of the region, inside the ellipse x^2 + 4y^2 = 4$x^{2} + 4y^{2} = 4$ and outside the region bounded by the curves y = |x| - 1$y = |x| - 1$ and y = 1 - |x|$y = 1 - |x|$ , is:
A.2(pi - 1)$2(\pi - 1)$
B.2pi - frac12$2\pi - \frac{1}{2}$
C.3(pi - 1)$3(\pi - 1)$
D.2pi - 1$2\pi - 1$
Solution & Explanation
### Related Formula
Area of an ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by:
textArea = pi a b$$\text{Area} = \pi a b$$
Area of a rhombus bounded by |x| + |y| = a$|x| + |y| = a$ is 2a^2$2a^2$.
### Core Logic
The given curves form a bounded geometric area.
Ellipse: x^2 + 4y^2 = 4 Rightarrow fracx^24 + fracy^21 = 1$x^2 + 4y^2 = 4 \Rightarrow \frac{x^2}{4} + \frac{y^2}{1} = 1$.
Here, a = 2$a = 2$, b = 1$b = 1$.
The region to be excluded is bounded by y = |x| - 1$y = |x| - 1$ and y = 1 - |x|$y = 1 - |x|$, which rearranges to |x| + |y| = 1$|x| + |y| = 1$. This forms a square/rhombus centered at the origin with vertices at (1, 0), (0, 1), (-1, 0), (0, -1)$(1, 0), (0, 1), (-1, 0), (0, -1)$.
### Step 1: Calculate Total and Excluded Areas
Total Area of the Ellipse:
textArea = pi (2)(1) = 2pi$$\text{Area} = \pi (2)(1) = 2\pi$$
Excluded Area (Rhombus |x| + |y| = 1$|x| + |y| = 1$):
The rhombus consists of 4 identical right-angled triangles in each quadrant.
Area of one triangle = frac12 times textbase times textheight = frac12 times 1 times 1 = frac12$\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.
Total excluded area = 4 times frac12 = 2$4 \times \frac{1}{2} = 2$.
### Step 2: Calculate Required Area
Ellipse and Modulus area diagram for Q2 - JEE Main 2026 Morning
Required Area = Area of ellipse -$-$ Shaded Area
= 2pi - 2$= 2\pi - 2$= 2(pi - 1)$= 2(\pi - 1)$
### Pattern Recognition
Transform absolute value equations y = pm(|x| - a)$y = \pm(|x| - a)$ into |x| + |y| = a$|x| + |y| = a$ to instantly recognize a standard rhombus, allowing direct geometry formulas instead of integration.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Area Under Curves
Class 11 Maths: Conic Sections
Keywords:#area of the region#JEE Main 2026 Morning Q2#Area Under Curves JEE Main 2026#Area Bounded by Ellipse and Modulus Functions JEE Main 2026
More Area Under Curves Previous-Year Questions — Page 4
Q28jee_main_2024_29_jan_morningArea Under Curves
The area (in sq. units) of the part of circle x^2+y^2=169$x^2+y^2=169$ which is below the line 5x-y=13$5x-y=13$ is fracpialpha2beta-frac652+fracalphabetasin^-1(frac1213)$\frac{\pi\alpha}{2\beta}-\frac{65}{2}+\frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$ where alpha,beta$\alpha,\beta$ are coprime numbers. Then alpha+beta$\alpha+\beta$ is equal to
Numerical Answer.Answer: 171 to 171
Solution
### Related Formula
textStandard Integral: int sqrta^2-y^2 dy = fracy2sqrta^2-y^2 + fraca^22sin^-1left(fracyaright) + C$$\text{Standard Integral: } \int \sqrt{a^2-y^2} dy = \frac{y}{2}\sqrt{a^2-y^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{y}{a}\right) + C$$textArea of right triangle = frac12 times textbase times textheight$$\text{Area of right triangle } = \frac{1}{2} \times \text{base} \times \text{height}$$
### Core Logic
First, find the points of intersection between the circle x^2+y^2=169$x^2+y^2=169$ and the line 5x-y=13 Rightarrow y = 5x-13$5x-y=13 \Rightarrow y = 5x-13$.
Substitute y$y$ into the circle equation:
x^2 + (5x-13)^2 = 169$$x^2 + (5x-13)^2 = 169$$x^2 + 25x^2 - 130x + 169 = 169$$x^2 + 25x^2 - 130x + 169 = 169$$26x^2 - 130x = 0 Rightarrow 26x(x - 5) = 0$$26x^2 - 130x = 0 \Rightarrow 26x(x - 5) = 0$$
The solutions are x=0$x=0$ and x=5$x=5$.
When x=0, y=-13$x=0, y=-13$. Point is (0, -13)$(0, -13)$.
When x=5, y=12$x=5, y=12$. Point is (5, 12)$(5, 12)$.
The required area is bounded below the line x = fracy+135$x = \frac{y+13}{5}$ and above the right-hand boundary of the circle x = sqrt169-y^2$x = \sqrt{169-y^2}$ across the y-axis boundaries [-13, 12]$[-13, 12]$.
Area Under Curves
### Step 1: Setup Area Integral
Integrate with respect to y$y$ (from left to right curves, bounded horizontally):
Area = int_-13^12 left( sqrt169-y^2 - fracy+135 right) dy$$Area = \int_{-13}^{12} \left( \sqrt{169-y^2} - \frac{y+13}{5} \right) dy$$
Split the integral into two parts:
Part A (Circle): int_-13^12 sqrt169-y^2 dy$\int_{-13}^{12} \sqrt{169-y^2} dy$
Part B (Line): int_-13^12 fracy+135 dy$\int_{-13}^{12} \frac{y+13}{5} dy$
### Step 2: Evaluate Integrals
Part A (Circle Integral):
= left[ fracy2sqrt169-y^2 + frac1692sin^-1left(fracy13right) right]_-13^12$$= \left[ \frac{y}{2}\sqrt{169-y^2} + \frac{169}{2}\sin^{-1}\left(\frac{y}{13}\right) \right]_{-13}^{12}$$
Evaluate at upper limit 12:
= frac122sqrt169-144 + frac1692sin^-1left(frac1213right) = 6(5) + frac1692sin^-1left(frac1213right) = 30 + frac1692sin^-1left(frac1213right)$$= \frac{12}{2}\sqrt{169-144} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right) = 6(5) + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right) = 30 + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$$
Evaluate at lower limit -13:
= 0 + frac1692sin^-1(-1) = -frac169pi4$$= 0 + \frac{169}{2}\sin^{-1}(-1) = -\frac{169\pi}{4}$$
Value of Part A = 30 + frac169pi4 + frac1692sin^-1left(frac1213right)$30 + \frac{169\pi}{4} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$
Part B (Line Integral - matches the area of the bounded triangle geometric region):
= frac110 left[ (y+13)^2 right]_-13^12$$= \frac{1}{10} \left[ (y+13)^2 \right]_{-13}^{12}$$= frac110(12+13)^2 - 0 = frac25^210 = frac62510 = frac1252 = 62.5$$= \frac{1}{10}(12+13)^2 - 0 = \frac{25^2}{10} = \frac{625}{10} = \frac{125}{2} = 62.5$$
### Step 3: Map to Requested Format
Subtract Part B from Part A:
Area = frac169pi4 + 30 - frac1252 + frac1692sin^-1left(frac1213right)$$Area = \frac{169\pi}{4} + 30 - \frac{125}{2} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$$Area = frac169pi4 - frac652 + frac1692sin^-1left(frac1213right)$$Area = \frac{169\pi}{4} - \frac{65}{2} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right)$$
Comparing this exactly with the given format fracpialpha2beta - frac652 + fracalphabetasin^-1(frac1213)$\frac{\pi\alpha}{2\beta} - \frac{65}{2} + \frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$:
We see that fracalphabeta = frac1692$\frac{\alpha}{\beta} = \frac{169}{2}$.
Since 169 and 2 are coprime, alpha = 169$\alpha = 169$ and beta = 2$\beta = 2$.
Calculate alpha + beta$\alpha + \beta$:
169 + 2 = 171$169 + 2 = 171$
### Pattern Recognition
When evaluating line integrals forming a triangle with horizontal bounds, bypass algebraic integration and visually calculate frac12 cdot b cdot h$\frac{1}{2} \cdot b \cdot h$. Here, base=25 along y-axis, height=5 along x-axis, area = 125/2$125/2$. Instantly saves integration time.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Application of Integrals
Class 11 Mathematics: Straight Lines
Q8jee_main_2024_30_jan_morningArea under Curves
The area (in square units) of the region bounded by the parabolay^2 = 4(x - 2)$y^2 = 4(x - 2)$ and the line y = 2x - 8$y = 2x - 8$
A.8$8$
B.9$9$
C.6$6$
D.7$7$
Solution
### Related Formula
textArea = int_y_1^y_2 (x_R - x_L) dy$$\text{Area} = \int_{y_1}^{y_2} (x_R - x_L) dy$$
### 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: y^2 = 4X Rightarrow X = fracy^24$y^2 = 4X \Rightarrow X = \frac{y^2}{4}$
Line: y = 2(X + 2) - 8 Rightarrow y = 2X - 4 Rightarrow X = fracy + 42$y = 2(X + 2) - 8 \Rightarrow y = 2X - 4 \Rightarrow X = \frac{y + 4}{2}$
### Step 1: Finding points of intersection
Set the X$X$ values equal to find intersection points in terms of y$y$:
fracy^24 = fracy + 42$$\frac{y^2}{4} = \frac{y + 4}{2}$$y^2 = 2y + 8$y^2 = 2y + 8$y^2 - 2y - 8 = 0$$y^2 - 2y - 8 = 0$$(y - 4)(y + 2) = 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_-2^4 left( x_R - x_L right) dy$$A = \int_{-2}^{4} \left( x_R - x_L \right) dy$$A = int_-2^4 left( fracy + 42 - fracy^24 right) dy$$A = \int_{-2}^{4} \left( \frac{y + 4}{2} - \frac{y^2}{4} \right) dy$$A = left[ fracy^24 + 2y - fracy^312 right]_-2^4$$A = \left[ \frac{y^2}{4} + 2y - \frac{y^3}{12} \right]_{-2}^{4}$$
Upper limit (y=4)$(y=4)$:
frac164 + 8 - frac6412 = 4 + 8 - frac163 = 12 - frac163 = frac203$\frac{16}{4} + 8 - \frac{64}{12} = 4 + 8 - \frac{16}{3} = 12 - \frac{16}{3} = \frac{20}{3}$
Lower limit (y=-2)$(y=-2)$:
frac44 - 4 - frac-812 = 1 - 4 + frac23 = -3 + frac23 = -frac73$\frac{4}{4} - 4 - \frac{-8}{12} = 1 - 4 + \frac{2}{3} = -3 + \frac{2}{3} = -\frac{7}{3}$A = frac203 - left(-frac73right) = frac273 = 9$$A = \frac{20}{3} - \left(-\frac{7}{3}\right) = \frac{27}{3} = 9$$
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Application of Integrals
Q11jee_main_2024_31_jan_eveningArea Under Curves
The area of the region enclosed by the parabola y = 4x - x^2$y = 4x - x^2$ and 3y = (x - 4)^2$3y = (x - 4)^2$ is equal to
Q5jee_main_2024_31_jan_morningArea bounded by Parabolas and Inequalities
The area of the region left\(x,y): y^2 le 4x, x < 4, fracxy(x - 1)(x - 2)(x - 3)(x - 4) > 0, x neq 3right\$\left\{(x,y): y^2 \le 4x, x < 4, \frac{xy(x - 1)(x - 2)}{(x - 3)(x - 4)} > 0, x \neq 3\right\}$ is
A.frac163$$\frac{16}{3}$$
B.frac643$$\frac{64}{3}$$
C.frac83$$\frac{8}{3}$$
D.frac323$$\frac{32}{3}$$
Solution
### Core Logic
Given y^2 le 4x$y^2 \le 4x$ and x < 4$x < 4$.
Analyze the inequality fracxy(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 fracx(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) cup (2, 3)$x \in (0, 1) \cup (2, 3)$.
### Step 2: Case II (y < 0)
If y < 0$y < 0$, the inequality reduces to fracx(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) cup (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 int_0^4 sqrtx dx = 2 cdot frac23[x^3/2]_0^4 = frac43 cdot 8 = frac323$= 2 \int_{0}^{4} \sqrt{x} dx = 2 \cdot \frac{2}{3}[x^{3/2}]_{0}^{4} = \frac{4}{3} \cdot 8 = \frac{32}{3}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Area Under Curves
Class 11 Maths: Linear Inequalities
More Area Under Curves Questions — jee_main_2026_21_jan_morning
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