Let y = p(x) be the parabola passing through the points (-1, 0), (0, 1) and (1, 0). If the area of t

Q: Wait, why do we only consider the lower semicircle for the boundary?

Great observation! The circle is centered at $(-1, 1)$ with radius $1$. The equation $(x+1)^2 + (y-1)^2 = 1$ gives $y = 1 \pm \sqrt{1 - (x+1)^2}$. Since our region is defined by $y \le p(x)$ and we are looking at the intersection, the parabola sits above the lower arc of the circle in the interval $[-1, 0]$. We choose the negative sign to represent the lower boundary of the region.

Q: Can I solve this without integration by using geometry?

Absolutely! That is the mark of a true physicist. The integral $\int_{-1}^0 \sqrt{1 - (x+1)^2} dx$ represents the area of a quarter-circle of radius $1$. Recognizing this geometric shape saves time and reduces the risk of calculation errors during the exam.

Q: How do I know which points to use for the parabola?

The problem gives you three points: $(-1, 0)$, $(0, 1)$, and $(1, 0)$. Since a parabola is defined by $p(x) = ax^2 + bx + c$, you have three unknowns. Three points provide exactly three linear equations, which is the perfect amount of information to solve for $a, b,$ and $c$ uniquely.

Q: Why does the integral for the parabola part result in a negative value?

The integral $\int_{-1}^0 -x^2 dx$ is negative because the function $f(x) = -x^2$ lies below the x-axis in the interval $[-1, 0]$. However, when you subtract this from the upper boundary, the signs will align to give you a positive area. Always trust the algebraic process!

Q: Is there a trick to the final expression $12(\pi - 4A)$?

Yes! It is designed to cancel out the transcendental $\pi$ term. When you calculate $A = \frac{\pi}{4} - \frac{1}{3}$, notice how multiplying by $4$ makes the $\pi$ term disappear. This is a classic JEE technique to ensure the final answer is a clean integer.

Question

Let $y = p(x)$ be the parabola passing through the points $(-1, 0), (0, 1)$ and $(1, 0)$. If the area of the region $\{(x, y): (x + 1)^2 + (y - 1)^2 \le 1, y \le p(x)\}$ is $A$, then $12(\pi - 4A)$ is equal to ________ .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we aren't just solving a problem; we are choreographing a dance between two fundamental shapes: the parabola and the circle.

Imagine yourself standing on a coordinate plane. You have a parabola, a graceful arc that kisses the x-axis at $(-1, 0)$ and $(1, 0)$ and peaks at $(0, 1)$. Then, you have a circle, a perfect, symmetric loop centered at $(-1, 1)$ with a radius of $1$.

### Defining the Parabola

Every great journey starts with a definition. We assume our parabola takes the form $p(x) = ax^2 + bx + c$.

By plugging in our three known points, we create a system of equations. Since the parabola passes through $(0, 1)$, we immediately see that $c = 1$.

Substituting the other two points, we find that $a + b + 1 = 0$ and $a - b + 1 = 0$. Solving these, we find $b = 0$ and $a = -1$. Our parabola is revealed:

$$p(x) = 1 - x^2$$

It is a simple, elegant downward-opening curve.

### The Intersection

Now, we look at the circle: $(x+1)^2 + (y-1)^2 = 1$. To find where these two shapes meet, we substitute our parabola $y = 1 - x^2$ into the circle's equation.

This leads us to $(x+1)^2 + ((1 - x^2) - 1)^2 = 1$, which simplifies beautifully to $(x+1)^2 + x^4 = 1$. Expanding this, we get $x^4 + x^2 + 2x = 0$.

Factoring out $x$, we find $x(x^3 + x + 2) = 0$. The real roots are $x = 0$ and $x = -1$. These are our boundaries. We are looking at the slice of the plane between $x = -1$ and $x = 0$.

### The Integral of Area

To find the area $A$, we integrate the difference between the upper boundary (the parabola) and the lower boundary (the lower arc of the circle). The lower arc is given by $y = 1 - \sqrt{1 - (x+1)^2}$.

Thus, our integral becomes:

$$A = \int_{-1}^{0} [ (1 - x^2) - (1 - \sqrt{1 - (x+1)^2}) ] dx$$

Distributing the negative sign, we get:

$$A = \int_{-1}^{0} [ -x^2 + \sqrt{1 - (x+1)^2} ] dx$$

This is where the magic happens. We split this into two parts. The first part, $\int_{-1}^{0} -x^2 dx$, is a straightforward polynomial integration, yielding $-\frac{1}{3}$.

The second part, $\int_{-1}^{0} \sqrt{1 - (x+1)^2} dx$, is the area of a quarter-circle of radius $1$, which is simply $\frac{\pi}{4}$.

### The Grand Finale

We have our area: $A = \frac{\pi}{4} - \frac{1}{3}$. The problem asks us to evaluate $12(\pi - 4A)$.

Substituting our value for $A$:

$$12\left(\pi - 4\left(\frac{\pi}{4} - \frac{1}{3}ight)ight) = 12\left(\pi - \pi + \frac{4}{3}ight) = 12 	imes \frac{4}{3} = 16$$

Look at that! The $\pi$ terms vanish, leaving us with a clean, satisfying integer.

The final answer is **16**. This is the beauty of mathematics—no matter how complex the path, the truth is often simple and elegant.

Animated Solution for Mathematics - Definite Integration: Let $y = p (x)$ be the parabola passing through the points $(- 1, 0), (0, 1)$ and $(1, 0)$ . If the area of the region ${(x, y) : (x + 1)^{2} + (y - 1)^{2} \leq 1, y \leq p (x)}$ is $A$ , then $12 (π - 4 A)$ is equal to ________ .

Enter Numerical Value:

Visualized Solution

$D e f inin g t h e P a r ab o l a p (x)$

$F in d in g t h e C o n s t an t c$

$S e tt in g u pE q u a t i o n s f or aan d b$

$E q u a t i o n o f t h e P a r ab o l a$

$A na l y z in g t h e C i r c l e E q u a t i o n$

$D e f inin g t h e R e g i o n A$

$F in d in g I n t er sec t i o n P o in t s$

$S o l v in g f or x$

$V i s u a l i z in g R e g i o n A$

$S e tt in g u pt h e I n t e g r a l f or A r e a A$

$S im pl i f y in g t h e I n t e g r a l$

$E v a l u a t in g t h e P a r ab o l i c P a r t$

$E v a l u a t in g t h e C i r c u l a r P a r t$

$T o t a l A r e a A$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Defining the Parabola

The Intersection

The Integral of Area

The Grand Finale

Similar Questions

The area of the region enclosed by the parabola $(y - 2)^{2} = x - 1$ , the line $x - 2 y + 4 = 0$ and the positive coordinate axes is_________.

The area of the region described by $A = {(x, y) : x^{2} + y^{2} \leq 1 and y^{2} \leq 1 - x}$ is:

If the area bounded by the curve $2 y^{2} = 3 x$ , lines $x + y = 3, y = 0$ and outside the circle $(x - 3)^{2} + y^{2} = 2$ is $A$ , then $4 (π + 4 A)$ is equal to _________.

Let the area of the region ${(x, y) : ∣2 x - 1∣ \leq y \leq ∣ x^{2} - x ∣, 0 \leq x \leq 1}$ be $A$ . Then $(6 A + 11)^{2}$ is equal to ______.

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The area (in sq. units) of the region enclosed between the parabola $y^{2} = 2 x$ and the line $x + y = 4$ is ____.

The area of the region $x, y : x^{2} \leq y \leq ∣ x^{2} - 4∣, y \geq 1$ is

The area of the region $A = {(x, y) : 0 \leq y \leq x^{2} + 1 and - 1 \leq x \leq 1}$ in sq. units, is :

The area (in sq. units) of the region bounded by the curves $x^{2} + 2 y - 1 = 0$ , $y^{2} + 4 x - 4 = 0$ and $y^{2} - 4 x - 4 = 0$ , in the upper half plane is ____.

The area of the region $A = {(x, y) : 4 x^{2} + y^{2} \leq 8 and y^{2} \leq 4 x}$ is:

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DefiningtheParabolap(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingtheConstantc

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SettingupEquationsforaandb

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EquationoftheParabola

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AnalyzingtheCircleEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DefiningtheRegionA

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingIntersectionPoints

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solvingforx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VisualizingRegionA

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SettinguptheIntegralforAreaA

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyingtheIntegral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EvaluatingtheParabolicPart

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EvaluatingtheCircularPart

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TotalAreaA

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalCalculation