For a 0, let the curves C1: y² = ax and C2: x² = ay intersect at origin O and a point P. Let the l

Q: Why is the area of $\Delta OQR$ simply $\frac{b^2}{2}$? I thought I needed coordinates for all vertices.

Great question! You are absolutely right that you need coordinates. The vertices are $O(0,0)$, $Q(b,b)$, and $R(b,0)$. Since $O$ is at the origin and $R$ is on the $x$-axis, the base is the segment $OR$ with length $b$. The height is the vertical distance from $R$ to $Q$, which is the $y$-coordinate of $Q$, also $b$. Thus, $\frac{1}{2} \times b \times b = \frac{b^2}{2}$. It is elegant, isn't it?

Q: Why did we square the equation $4a^{3/2} = a^3 + 2$? Isn't there a risk of introducing extraneous roots?

You have a sharp eye for detail! Yes, squaring can introduce extraneous roots. However, in the context of JEE problems, we are looking for the equation that *satisfies* the condition. Since $a > 0$, the squaring process is a valid algebraic maneuver to transform the transcendental-looking equation into a polynomial one. We are not solving for $a$ here, but finding the polynomial relation that $a$ must satisfy.

Q: Can I use symmetry to solve the area integral faster?

Absolutely! The curves $y^2 = ax$ and $x^2 = ay$ are reflections of each other across the line $y=x$. The area under $y^2=ax$ from $0$ to $a$ is the same as the area under $x^2=ay$ from $0$ to $a$. You could calculate the area under one curve and subtract the area of the triangle formed by the chord, but the standard integration method $\int (y_{upper} - y_{lower}) dx$ is robust and less prone to sign errors. Stick to the standard method unless you are very confident in your geometric visualization.

Question

For $a > 0$, let the curves $C_1: y^2 = ax$ and $C_2: x^2 = ay$ intersect at origin $O$ and a point $P$. Let the line $x = b$ $(0 < b < a)$ intersect the chord $OP$ and the $x$-axis at points $Q$ and $R$, respectively. If the line $x = b$ bisects the area bounded by the curves, $C_1$ and $C_2$, and the area of $\Delta OQR = \frac{1}{2}$, then '$a$' satisfies the equation :

Options:

(A) $a^6 - 12a^3 + 4 = 0$

(B) $a^6 - 12a^3 - 4 = 0$

(C) $a^6 + 6a^3 - 4 = 0$

(D) $a^6 - 6a^3 + 4 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before a coordinate plane, watching two curves emerge from the origin. We have $C_1: y^2 = ax$, a parabola stretching its arms to the right, and $C_2: x^2 = ay$, a parabola reaching upwards toward the sky.

These curves are reflections of each other, dancing around the line $y=x$. Our journey begins by finding where they meet.

By substituting $y = \frac{x^2}{a}$ into $y^2 = ax$, we find $x^4 = a^3x$. This yields $x=0$ and $x=a$. Thus, our intersection point $P$ is at $(a, a)$. This point is the anchor of our problem.

### The Chord and the Triangle

Now, consider the chord $OP$. Since it connects $(0,0)$ and $(a,a)$, its equation is simply $y=x$. This is a beautiful, clean line that cuts through the heart of our region.

We introduce a vertical line $x=b$. It slices through our chord at $Q(b,b)$ and hits the $x$-axis at $R(b,0)$. We are told the area of $\Delta OQR$ is $\frac{1}{2}$.

With a base of $b$ and a height of $b$, the area is $\frac{1}{2}b^2$. Equating this to $\frac{1}{2}$, we find $b=1$. The geometry has yielded its secret; the line of bisection is fixed at $x=1$.

### The Calculus of Bisection

Now, we must calculate the total area trapped between the curves. We integrate the difference between the upper curve and the lower curve:

$$A = \int_{0}^{a} \left( \sqrt{ax} - \frac{x^2}{a} \right) dx$$

Performing the integration, we get:

$$\left[ \frac{2}{3}\sqrt{a}x^{3/2} - \frac{x^3}{3a} \right]_0^a = \frac{a^2}{3}$$

The problem states that the line $x=1$ bisects this area. Therefore, the integral from $0$ to $1$ must equal half of the total area, which is $\frac{a^2}{6}$.

### The Final Algebraic Symphony

We set up our final equation:

$$\int_{0}^{1} \left( \sqrt{ax} - \frac{x^2}{a} \right) dx = \frac{a^2}{6}$$

Evaluating this, we get:

$$\frac{2\sqrt{a}}{3} - \frac{1}{3a} = \frac{a^2}{6}$$

To clear the denominators, we multiply by $6a$, yielding $4a\sqrt{a} - 2 = a^3$, or $4a^{3/2} = a^3 + 2$. To eliminate the fractional power, we square both sides:

$$(4a^{3/2})^2 = (a^3 + 2)^2$$

This expands to $16a^3 = a^6 + 4a^3 + 4$. Rearranging gives us the final, elegant polynomial:

**$$a^6 - 12a^3 + 4 = 0$$**

We have traversed the landscape of geometry, integration, and algebra to arrive at the truth. This is the essence of JEE Advanced—not just solving, but understanding the beautiful interplay of mathematical concepts.

Select Answer:

Visualized Solution

Visualizing the Curves

Finding Point $P$

Equation of Chord $O P$

Defining Points $Q$ and $R$

Area of Triangle $O QR$

Total Bounded Area Setup

Calculating Total Area

Area Bisection Condition

Evaluating the Left Area Integral

Simplifying the Equation

Squaring to Remove Roots

Final Equation for $a$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Chord and the Triangle

The Calculus of Bisection

The Final Algebraic Symphony

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