The sum of squares of all possible values of k, for which area of the region bounded by the parabola

Q: Wait, why can't we just integrate with respect to x? It feels more natural.

That is a fantastic question! While integrating with respect to $x$ is standard, look at the equations: $2y^2 = kx$ implies $y = \pm \sqrt{\frac{kx}{2}}$. If you integrate with respect to $x$, you have to deal with square roots and split the integral into multiple parts because of the $\pm$ sign. By integrating with respect to $y$, we treat $x$ as a function of $y$, which gives us simple polynomials. In JEE Advanced, choosing the right variable of integration is half the battle won!

Q: Why do we accept $k = -2$ as a solution? Doesn't area have to be positive?

You are absolutely right that the area itself must be positive! However, the question asks for the sum of squares of all possible values of $k$. Even if $k$ is negative, the term $k^2$ in our area function $A(k) = \frac{2}{3} \frac{k^2}{(k^2+4)^2}$ ensures the area remains positive. Both $k=2$ and $k=-2$ yield the same maximum area. Always check the domain of your variables, but don't discard negative values unless the problem explicitly restricts $k$ to be positive!

Q: Is the AM-GM inequality always valid for any $k$?

The standard AM-GM inequality $\frac{a+b}{2} \ge \sqrt{ab}$ applies to positive real numbers. If $k$ is negative, we can look at the symmetry of the function $A(k)$. Since $A(k)$ depends on $k^2$, the behavior for negative $k$ is identical to positive $k$. We essentially apply AM-GM to $|k|$ and $|4/k|$, which leads us to the same critical values. It is a robust approach!

Question

The sum of squares of all possible values of $k$, for which area of the region bounded by the parabolas $2y^2 = kx$ and $ky^2 = 2(y - x)$ is maximum, is equal to :

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: A Journey into Parabolas and Optimization

Welcome, future engineer. Today, we are not just solving a math problem; we are embarking on a journey to uncover the hidden symmetry within coordinate geometry.

We are given two parabolas: $2y^2 = kx$ and $ky^2 = 2(y - x)$. At first glance, they might look like a tangled mess of variables, but I want you to pause and breathe. In the world of JEE Advanced, complexity is often just a mask for a deeper, simpler truth waiting to be revealed.

### Phase 1: Changing the Perspective

Most students instinctively try to solve for $y$ in terms of $x$. But look at the equations again. If we solve for $y$, we get square roots, and nobody wants to integrate those!

Instead, let us flip our perspective and express $x$ as a function of $y$. The first curve becomes:

$$x = \frac{2y^2}{k}$$

The second curve transforms into:

$$x = y - \frac{ky^2}{2}$$

Suddenly, the curves are just simple parabolas opening along the $x$-axis. By choosing the right perspective, we have already simplified the problem significantly.

### Phase 2: The Intersection

To find the region bounded by these curves, we need to know where they meet. We equate the two expressions for $x$:

$$\frac{2y^2}{k} = y - \frac{ky^2}{2}$$

Bringing everything to one side, we get:

$$y \left( \frac{2y}{k} + \frac{ky}{2} - 1 ight) = 0$$

This gives us our limits of integration: $y = 0$ and $y = \frac{2k}{k^2+4}$. These are the boundaries of our world. Everything happening between these two values is what we need to measure.

### Phase 3: The Integral of Discovery

Now, we set up the integral for the area $A = \int_{0}^{\frac{2k}{k^2+4}} (x_{	ext{right}} - x_{	ext{left}}) \, dy$. Substituting our expressions, we get:

$$A = \int_{0}^{\frac{2k}{k^2+4}} \left( (y - \frac{ky^2}{2}) - \frac{2y^2}{k} ight) \, dy$$

Grouping the $y^2$ terms, we have:

$$A = \int_{0}^{\frac{2k}{k^2+4}} \left( y - y^2 \left( \frac{k}{2} + \frac{2}{k} ight) ight) \, dy$$

Integrating term by term, we get:

$$\left[ \frac{y^2}{2} - \frac{y^3}{3} \left( \frac{k^2+4}{2k} ight) ight]_0^{\frac{2k}{k^2+4}}$$

When we plug in the upper limit, the terms simplify beautifully, leaving us with:

$$A = \frac{2}{3} \frac{k^2}{(k^2+4)^2}$$

### Phase 4: The Optimization

We are almost there. We need to maximize this area. Let us rewrite the area function as:

$$A = \frac{2}{3} \frac{1}{(k + \frac{4}{k})^2}$$

To make $A$ as large as possible, we must make the denominator $(k + \frac{4}{k})^2$ as small as possible. This is where the **AM-GM inequality** shines.

For any positive $k$, $k + \frac{4}{k} \ge 2\sqrt{k \cdot \frac{4}{k}} = 4$. The minimum value of the denominator is $4^2 = 16$. This minimum occurs when $k = \frac{4}{k}$, which means $k^2 = 4$. Thus, $k = 2$ or $k = -2$.

### The Final Victory

The question asks for the sum of the squares of all possible values of $k$. We found $k = 2$ and $k = -2$.

Squaring them gives $4$ and $4$. Adding them together, we get **8**.

We have navigated the geometry, conquered the calculus, and utilized the power of inequalities to reach the finish line. Remember, the math is not just about the answer; it is about the elegance of the path taken.

Animated Solution for Mathematics - Definite Integration: The sum of squares of all possible values of $k$ , for which area of the region bounded by the parabolas $2 y^{2} = k x$ and $k y^{2} = 2 (y - x)$ is maximum, is equal to :

Enter Numerical Value:

Visualized Solution

Identify the Parabolas

Equating the Curves

Finding Intersection Points

Setting up the Integral

The Area Expression

Integrating with respect to $y$

Evaluating the Definite Integral

Simplifying the Area Function

Preparing to Maximize

Applying AM-GM Inequality

Condition for Maximum Area

Sum of Squares of $k$

The Sigma Insight: Area Bounded by Curves

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

The Final Victory

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