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.
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: