The area of the smaller region enclosed by the curves y² = 8x + 4 and x² + y² + 4√3x - 4 = 0 is equa

Q: Why did we reject the intersection point $x = -(8 + 4\sqrt{3})$?

Great catch! When we solve the quadratic equation for intersection, we get two potential $x$ values. However, we must check if these correspond to real $y$ values on the parabola. Substituting $x = -(8 + 4\sqrt{3})$ into $y^2 = 8x + 4$ yields a negative value for $y^2$. Since the square of a real number cannot be negative, this point is purely algebraic and does not exist on the graph. Always verify your intersection points!

Q: Why integrate with respect to $y$ instead of $x$?

Integrating with respect to $x$ would force you to split the region into multiple parts because the upper and lower halves of the circle and parabola would create complex boundary conditions. By using horizontal strips (integrating with respect to $y$), the boundaries are consistent functions across the entire interval, making the calculus much cleaner and less prone to error.

Q: Is there a shortcut for the integral of $\sqrt{16 - y^2}$?

While you can use trigonometric substitution ($y = 4\sin\theta$), memorizing the standard form $\int \sqrt{a^2 - y^2} dy = \frac{y}{2}\sqrt{a^2 - y^2} + \frac{a^2}{2}\sin^{-1}(\frac{y}{a})$ is a massive time-saver for JEE Advanced. It turns a multi-step substitution problem into a simple evaluation.

Question

The area of the smaller region enclosed by the curves $y^2 = 8x + 4$ and $x^2 + y^2 + 4\sqrt{3}x - 4 = 0$ is equal to

Options:

(A) $\frac{1}{3}(2 - 12\sqrt{3} + 8\pi)$

(B) $\frac{1}{3}(2 - 12\sqrt{3} + 6\pi)$

(C) $\frac{1}{3}(4 - 12\sqrt{3} + 8\pi)$

(D) $\frac{1}{3}(4 - 12\sqrt{3} + 6\pi)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on the Cartesian plane, looking at two distinct mathematical entities: a parabola and a circle. The parabola, $y^2 = 8x + 4$, is a wide, sweeping curve that opens its arms to the right, anchored at a vertex of $(-\frac{1}{2}, 0)$.

The circle, $x^2 + y^2 + 4\sqrt{3}x - 4 = 0$, is a perfect, closed loop. It is centered at $(-2\sqrt{3}, 0)$ with a radius of $4$. Our mission is to find the area of the smaller region where these two paths collide.

### Phase 1: The Collision

To find the area, we must first determine where these curves meet. We substitute the parabola's $y^2$ expression directly into the circle's equation:

$$x^2 + (8x + 4) + 4\sqrt{3}x - 4 = 0$$

Notice how the constants $4$ and $-4$ vanish, leaving us with the simplified quadratic:

$$x^2 + (8 + 4\sqrt{3})x = 0$$

This factors beautifully into $x(x + 8 + 4\sqrt{3}) = 0$. We find two intersection candidates: $x = 0$ and $x = -(8 + 4\sqrt{3})$.

The second value is a ghost—it yields no real $y$ values. Thus, the curves shake hands only at $x = 0$, which corresponds to $y = 2$ and $y = -2$.

### Phase 2: Choosing the Path of Least Resistance

Now, we must calculate the area. Integrating with respect to $x$ would result in a complex split integral, so we choose horizontal strips instead. By integrating with respect to $y$, we define our boundaries as functions of $y$.

The right boundary is the circle:

$$x_C = \sqrt{16 - y^2} - 2\sqrt{3}$$

The left boundary is the parabola:

$$x_P = \frac{y^2 - 4}{8}$$

The area is the integral of the right curve minus the left curve.

### Phase 3: The Beauty of Symmetry

Because the region is perfectly symmetric about the x-axis, we can calculate the area from $0$ to $2$ and double it. Our integral becomes:

$$A = 2 \int_{0}^{2} \left[ \left( \sqrt{16 - y^2} - 2\sqrt{3} ight) - \left( \frac{y^2 - 4}{8} ight) ight] dy$$

We break this into three manageable pieces:

$$I_1 = \int_{0}^{2} \sqrt{16 - y^2} \, dy, \quad I_2 = \int_{0}^{2} 2\sqrt{3} \, dy, \quad I_3 = \int_{0}^{2} \frac{y^2 - 4}{8} \, dy$$

### Phase 4: The Final Calculation

$I_1$ is the classic integral of a circular segment, evaluating to $2\sqrt{3} + \frac{4\pi}{3}$. $I_2$ is a simple constant integral, giving $4\sqrt{3}$. $I_3$ is a polynomial integral, resulting in $-\frac{2}{3}$.

When we combine these, we get:

$$A = 2 \left[ \left( 2\sqrt{3} + \frac{4\pi}{3} ight) - 4\sqrt{3} - \left( -\frac{2}{3} ight) ight]$$

Simplifying the terms inside the bracket, we obtain:

$$A = 2 \left[ \frac{4\pi}{3} - 2\sqrt{3} + \frac{2}{3} ight]$$

Factoring out $\frac{1}{3}$, we arrive at the final, elegant result:

**$$	ext{Area} = \frac{1}{3}(8\pi - 12\sqrt{3} + 4)$$**

This matches our target, proving that even the most complex problems yield to a structured, patient approach. You have mastered the geometry and the calculus—well done!

Animated Solution for Mathematics - Definite Integration: The area of the smaller region enclosed by the curves $y^{2} = 8 x + 4$ and $x^{2} + y^{2} + 43 x - 4 = 0$ is equal to

Select Answer:

Visualized Solution

Analyze the Parabola

Analyze the Circle

Find Intersection Points

Determine Valid $y$ Values

Visualize the Region

Set up the Integral

Integrate the Circle Part

Integrate the Remaining Parts

Combine and Calculate

Final Result

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

The area of the region enclosed by the curves $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$ is:

The area of the region enclosed between the circles $x^{2} + y^{2} = 4$ and $x^{2} + (y - 2)^{2} = 4$ is:

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

Let $A_{1}$ be the bounded area enclosed by the curves $y = x^{2} + 2, x + y = 8$ and $y$ -axis that lies in the first quadrant. Let $A_{2}$ be the bounded area enclosed by the curves $y = x^{2} + 2, y^{2} = x, x = 2$ , and $y$ -axis that lies in the first quadrant. Then $A_{1} - A_{2}$ is equal to

The area (in sq. units) of the region ${(x, y) : y^{2} \geq 2 x and x^{2} + y^{2} \leq 4 x, x \geq 0, y \geq 0}$ is:

The area of the region bounded by $y^{2} = 8 x$ and $y^{2} = 16 (3 - x)$ is equal to :-

The area of the region enclosed by the parabola $y = 4 x - x^{2}$ and $3 y = (x - 4)^{2}$ is equal to

The area of the region in the first quadrant inside the circle $x^{2} + y^{2} = 8$ and outside the parabola $y^{2} = 2 x$ is equal to :

The area of the region, inside the circle $(x - 23)^{2} + y^{2} = 12$ and outside the parabola $y^{2} = 23 x$ is:

The area of the region, inside the ellipse $x^{2} + 4 y^{2} = 4$ and outside the region bounded by the curves $y = ∣ x ∣ - 1$ and $y = 1 - ∣ x ∣$ , is :

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the smaller region enclosed by the curves y2=8x+4 and x2+y2+43​x−4=0 is equal to