What is the locus of points, the difference of whose distances from two points being constant?

Options

A. Pair of straight lines
B. An ellipse
C. A hyperbola
D. A parabola

Answer

To find the locus of points where the difference of the distances from two fixed points (let's call them F1 and F2) is constant, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Fixed Points**: Let F1 and F2 be two fixed points in the coordinate plane. For simplicity, we can place F1 at (−c, 0) and F2 at (c, 0), where c is a positive constant.

2. **Distance from a Point to Fixed Points**: Let P(x, y) be any point in the plane. The distance from P to F1 is given by:
\[ d_1 = \sqrt{(x + c)^2 + y^2} \]
and the distance from P to F2 is given by:
\[ d_2 = \sqrt{(x - c)^2 + y^2} \]

3. **Set Up the Condition**: According to the problem, the difference of these distances is constant. Let's denote this constant by k. Therefore, we have:
\[ |d_2 - d_1| = k \]

4. **Square Both Sides**: To eliminate the square roots, we can square both sides of the equation. However, we need to consider both cases of the absolute value:
- Case 1: \(d_2 - d_1 = k\)
- Case 2: \(d_1 - d_2 = k\)

For Case 1:
\[ d_2 - d_1 = k \implies \sqrt{(x - c)^2 + y^2} - \sqrt{(x + c)^2 + y^2} = k \]

For Case 2:
\[ d_1 - d_2 = k \implies \sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = k \]

5. **Rearranging and Squaring**: We can rearrange the first case and square both sides:
\[ \sqrt{(x - c)^2 + y^2} = k + \sqrt{(x + c)^2 + y^2} \]
Squaring both sides gives:
\[ (x - c)^2 + y^2 = (k + \sqrt{(x + c)^2 + y^2})^2 \]

Expanding this will lead to a quadratic equation in terms of x and y, which will ultimately describe a hyperbola.

6. **Conclusion**: After simplifying the resulting equation, we find that the locus of points P(x, y) such that the difference of the distances to the two fixed points F1 and F2 is constant is a hyperbola.

### Final Answer:
The locus of points, the difference of whose distances from two fixed points is constant, is a **hyperbola**.