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

Q: Why can I just double the integral from $0$ to $3$ for $A_1$? Is that always allowed?

That is a fantastic question! You can only use that shortcut when the integrand is an 'even function' and the limits of integration are symmetric about the origin (i.e., from $-a$ to $a$). In our case, $27 - 3x^2$ is indeed an even function because replacing $x$ with $-x$ leaves the expression unchanged. Since the region is perfectly symmetric across the y-axis, calculating the area on the right side and doubling it is mathematically sound and saves you precious time during the exam.

Q: How do I know which function is the 'upper' one in the integral? What if I get it backward?

If you get the order wrong, you will simply end up with a negative area. Area must always be positive. A reliable way to check is to pick a test point between your limits of integration. For $A_1$, pick $x=0$. $P_2$ gives $y=27$, and $P_1$ gives $y=0$. Since $27 > 0$, $P_2$ is clearly the upper curve. Always perform this quick 'sanity check' before you start integrating to ensure your result is positive.

Q: Is there a faster way to find the area between a parabola and a line without doing the full integration?

Yes, there is! For a parabola $y = ax^2$ and a line $y = mx + c$, the area enclosed is given by $\frac{(x_2 - x_1)^3}{6} \cdot |a|$, where $x_1$ and $x_2$ are the intersection points. While this is a powerful tool, I strongly encourage you to master the integration method first. In JEE Advanced, they sometimes tweak the curves to be non-standard, and the integration method is the universal key that will never fail you.

Question

Let $P_1 : y = 4x^2$ and $P_2 : y = x^2 + 27$ be two parabolas. If the area of the bounded region enclosed between $P_1$ and $P_2$ is six times the area of the bounded region enclosed between the line $y = \alpha x, \alpha > 0$ and $P_1$, then $\alpha$ is equal to :

Options:

(A) 12

(B) 6

(C) 15

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are exploring the architecture of curves.

We have two parabolas, $P_1: y = 4x^2$ and $P_2: y = x^2 + 27$. Imagine them as two distinct paths in a coordinate plane.

$P_1$ is the tight, focused parabola, hugging the origin, while $P_2$ is the elevated, broader sibling, resting its vertex at $y = 27$. Our mission is to understand the space trapped between them and then see how a simple line, $y = \alpha x$, cuts through this geometry.

### The Intersection of Paths

Before we can calculate the area $A_1$ trapped between $P_1$ and $P_2$, we must find where they meet. This is the 'anchor' of our problem.

We set the equations equal to each other:

$$4x^2 = x^2 + 27$$

Subtracting $x^2$ from both sides, we arrive at $3x^2 = 27$, which simplifies beautifully to $x^2 = 9$.

This tells us that our parabolas intersect at $x = -3$ and $x = 3$. These are our boundaries. We are looking at the region where $x$ ranges from $-3$ to $3$.

### Calculating the First Enclosure ($A_1$)

Now, we define the area. The area between two curves is the integral of the 'upper' curve minus the 'lower' curve. By looking at our sketch, we see that $P_2$ sits above $P_1$ in this interval.

Thus, our integral is:

$$A_1 = \int_{-3}^{3} (P_2 - P_1) dx = \int_{-3}^{3} (x^2 + 27 - 4x^2) dx$$

Simplifying the integrand, we get $\int_{-3}^{3} (27 - 3x^2) dx$. Here is where we apply a bit of mathematical elegance.

Since the function $f(x) = 27 - 3x^2$ is an even function, we can exploit the symmetry of the parabola. Instead of integrating across the entire range, we integrate from $0$ to $3$ and double the result:

$$A_1 = 2 \int_{0}^{3} (27 - 3x^2) dx$$

Performing the integration, we get $2 [27x - x^3]_0^3$. Plugging in our limits, we find $2(81 - 27) = 2(54) = 108$.

We have successfully quantified the first region. **$A_1 = 108$ square units.**

### The Intruder ($y = \alpha x$)

Now, the problem introduces a new player: the line $y = \alpha x$. This line passes through the origin and slices through $P_1$.

We need to find the area $A_2$ enclosed by this line and $P_1$. First, we find the intersection points:

$$4x^2 = \alpha x \Rightarrow x(4x - \alpha) = 0$$

This gives us $x = 0$ and $x = \frac{\alpha}{4}$. These are the limits for our second integral.

In this region, the line is above the parabola, so we integrate the line minus the parabola:

$$A_2 = \int_{0}^{\frac{\alpha}{4}} (\alpha x - 4x^2) dx$$

### The Final Synthesis

Let us evaluate $A_2$ with care. The integral of $\alpha x$ is $\frac{\alpha x^2}{2}$, and the integral of $4x^2$ is $\frac{4x^3}{3}$.

Evaluating from $0$ to $\frac{\alpha}{4}$:

$$A_2 = \left[ \frac{\alpha x^2}{2} - \frac{4x^3}{3} \right]_0^{\frac{\alpha}{4}}$$

Substituting the upper limit $\frac{\alpha}{4}$:

$$A_2 = \frac{\alpha}{2} \left( \frac{\alpha^2}{16} \right) - \frac{4}{3} \left( \frac{\alpha^3}{64} \right) = \frac{\alpha^3}{32} - \frac{\alpha^3}{48}$$

Finding a common denominator of $96$, we get $A_2 = \frac{3\alpha^3 - 2\alpha^3}{96} = \frac{\alpha^3}{96}$.

Finally, we use the condition given in the problem: $A_1 = 6 A_2$. Substituting our values:

$$108 = 6 \left( \frac{\alpha^3}{96} \right)$$

Simplifying the right side, $108 = \frac{\alpha^3}{16}$. Multiplying by $16$, we get $\alpha^3 = 1728$.

Taking the cube root, we find **$\alpha = 12$**.

Select Answer:

Visualized Solution

Visualizing $P_{1}$ and $P_{2}$

Finding Intersection Points

Setting up Area $A_{1}$

Evaluating Area $A_{1}$

Introducing Line $y = α x$

Intersection of $L$ and $P_{1}$

Setting up Area $A_{2}$

Evaluating Area $A_{2}$

Applying the Area Condition

Solving for $α^{3}$

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Intersection of Paths

Calculating the First Enclosure ( $A_{1}$ )

The Intruder ( $y = α x$ )

The Final Synthesis

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Select Answer:

Visualized Solution

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​

Finding Intersection Points

Setting up Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​

Evaluating Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​

Introducing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=αx

Intersection of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​

Setting up Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​

Evaluating Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​

Applying the Area Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α3

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Intersection of Paths

Calculating the First Enclosure ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​)

The Intruder ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=αx)