A point moves such that the sum of the squares of its distances from the two sides of length ‘a’ of a rectangle is twice the sum of the squares of its distances from the other two sides of length b. The locus of the point can be:

Options

A. a circle
B. an ellipse
C. a hyperbola
D. a pair of lines

Explanation

To solve the problem, we need to find the locus of a point \( P(h, k) \) that moves such that the sum of the squares of its distances from the two sides of length \( a \) of a rectangle is twice the sum of the squares of its distances from the other two sides of length \( b \).

### Step-by-Step Solution:

1. **Understanding the Rectangle**:
- Let's position the rectangle in the coordinate plane. The rectangle has vertices at \( (0, 0) \), \( (a, 0) \), \( (a, b) \), and \( (0, b) \).
- The sides of length \( a \) are along the lines \( y = 0 \) and \( y = b \).
- The sides of length \( b \) are along the lines \( x = 0 \) and \( x = a \).

2. **Distances from the Sides**:
- The distance from the point \( P(h, k) \) to the sides of length \( a \) (i.e., \( y = 0 \) and \( y = b \)) is:
- Distance to \( y = 0 \): \( k \)
- Distance to \( y = b \): \( b - k \)
- The sum of the squares of these distances is:
\[ k^2 + (b - k)^2 = k^2 + (b^2 - 2bk + k^2) = 2k^2 - 2bk + b^2 \]

3. **Distances from the Other Sides**:
- The distance from the point \( P(h, k) \) to the sides of length \( b \) (i.e., \( x = 0 \) and \( x = a \)) is:
- Distance to \( x = 0 \): \( h \)
- Distance to \( x = a \): \( a - h \)
- The sum of the squares of these distances is:
\[ h^2 + (a - h)^2 = h^2 + (a^2 - 2ah + h^2) = 2h^2 - 2ah + a^2 \]

4. **Setting Up the Equation**:
- According to the problem, the sum of the squares of the distances from the sides of length \( a \) is twice the sum of the squares of the distances from the sides of length \( b \):
\[ 2k^2 - 2bk + b^2 = 2(2h^2 - 2ah + a^2) \]
- Simplifying the right side:
\[ 2k^2 - 2bk + b^2 = 4h^2 - 4ah + 2a^2 \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ 2k^2 - 4h^2 - 2bk + 4ah + b^2 - 2a^2 = 0 \]
- Dividing the entire equation by 2:
\[ k^2 - 2h^2 - bk + 2ah + \frac{b^2 - 2a^2}{2} = 0 \]

6. **Identifying the Locus**:
- This is a quadratic equation in \( k \) and can be expressed in the standard form of a conic section.
- Rearranging gives:
\[ k^2 - 2h^2 - bk + (2ah + \frac{b^2 - 2a^2}{2}) = 0 \]
- The equation resembles the standard form of a hyperbola.

7. **Conclusion**:
- The locus of the point \( P(h, k) \) is a hyperbola.

### Final Answer:
The locus of the point can be a hyperbola.