Question

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

Answer 1

Given parabolas:
$2y^2 = kx \quad \text{and} \quad ky^2 = 2(y - x)$

Step 1: Simplifying the Equations
For the first parabola:
$y^2 = \frac{kx}{2}$
This represents a parabola opening towards the positive $x$-axis.
For the second parabola:
$ky^2 = 2y - 2x \implies y^2 = \frac{2y - 2x}{k}$
This represents a parabola whose orientation depends on the value of $k$.

Step 2: Finding the Intersection Points
To find the intersection points, we equate the two expressions for $y^2$:
$\frac{kx}{2} = \frac{2y - 2x}{k}$
Rearranging:
$k^2x = 4y - 4x$
Further simplification yields a relationship between $x$ and $y$ that can be analyzed to find the conditions on $k$ for maximizing the bounded area.

Step 3: Maximizing the Area
To maximize the area of the region bounded by the parabolas, we find the values of $k$ that lead to maximum enclosed regions. By analyzing the geometry of the parabolas and their orientations, we find that the optimal values of $k$ are:
$k = 2 \quad \text{and} \quad k = -2$

Step 4: Calculating the Sum of Squares of All Possible Values of $k$
The sum of squares of all possible values of $k$ is:
$k^2 + (-k)^2 = 2^2 + (-2)^2 = 4 + 4 = 8$

Conclusion: The sum of squares of all possible values of $k$ is 8.