Find the locus of a point such that the sum of its distances from the points (0, 2) and (0, -2) is 6.

Answer

To find the locus of a point such that the sum of its distances from the points (0, 2) and (0, -2) is 6, we can follow these steps:

### Step 1: Define the Point
Let the point be \( P(h, k) \).

### Step 2: Calculate the Distances
1. The distance from \( P(h, k) \) to \( (0, 2) \):
\[ d_1 = \sqrt{(h - 0)^2 + (k - 2)^2} = \sqrt{h^2 + (k - 2)^2} \]
Expanding this gives:
\[ d_1 = \sqrt{h^2 + k^2 - 4k + 4} \]

2. The distance from \( P(h, k) \) to \( (0, -2) \):
\[ d_2 = \sqrt{(h - 0)^2 + (k + 2)^2} = \sqrt{h^2 + (k + 2)^2} \]
Expanding this gives:
\[ d_2 = \sqrt{h^2 + k^2 + 4k + 4} \]

### Step 3: Set Up the Equation
According to the problem, the sum of these distances is equal to 6:
\[ d_1 + d_2 = 6 \]
Substituting the distances:
\[ \sqrt{h^2 + k^2 - 4k + 4} + \sqrt{h^2 + k^2 + 4k + 4} = 6 \]

### Step 4: Isolate One Square Root
Let:
\[ x = \sqrt{h^2 + k^2 - 4k + 4} \]
\[ y = \sqrt{h^2 + k^2 + 4k + 4} \]
Then we have:
\[ x + y = 6 \]
Isolating \( y \):
\[ y = 6 - x \]

### Step 5: Square Both Sides
Square both sides to eliminate the square roots:
\[ y^2 = (6 - x)^2 \]
Substituting \( y \):
\[ h^2 + k^2 + 4k + 4 = (6 - \sqrt{h^2 + k^2 - 4k + 4})^2 \]

### Step 6: Expand and Simplify
Expanding the right side:
\[ (6 - x)^2 = 36 - 12x + x^2 \]
Setting both sides equal:
\[ h^2 + k^2 + 4k + 4 = 36 - 12\sqrt{h^2 + k^2 - 4k + 4} + (h^2 + k^2 - 4k + 4) \]

### Step 7: Combine Like Terms
Combine like terms:
\[ h^2 + k^2 + 4k + 4 = h^2 + k^2 + 4 - 12\sqrt{h^2 + k^2 - 4k + 4} \]
This simplifies to:
\[ 4k = -12\sqrt{h^2 + k^2 - 4k + 4} \]

### Step 8: Isolate the Square Root
Isolate the square root:
\[ \sqrt{h^2 + k^2 - 4k + 4} = -\frac{1}{3}k \]

### Step 9: Square Again
Square both sides again:
\[ h^2 + k^2 - 4k + 4 = \frac{1}{9}k^2 \]

### Step 10: Rearranging
Rearranging gives:
\[ 9h^2 + 9k^2 - 36k + 36 = k^2 \]
\[ 9h^2 + 8k^2 - 36k + 36 = 0 \]

### Step 11: Final Form
This represents the equation of an ellipse:
\[ \frac{h^2}{5} + \frac{k^2}{9} = 1 \]
Replacing \( h \) with \( x \) and \( k \) with \( y \):
\[ \frac{x^2}{5} + \frac{y^2}{9} = 1 \]

### Conclusion
Thus, the locus of the point is an ellipse given by:
\[ \frac{x^2}{5} + \frac{y^2}{9} = 1 \]