A point P(x, y) moves so that the sum of its distances from the points S(4, 2) and `S^(')` (-2, 2) is 8. Find the equation of its locus and show that it is an ellipse.

Answer

To find the equation of the locus of the point P(x, y) such that the sum of its distances from the points S(4, 2) and S'(-2, 2) is 8, we can follow these steps:

### Step 1: Write the distance formula
The distance from point P(x, y) to point S(4, 2) is given by:
\[ d_1 = \sqrt{(x - 4)^2 + (y - 2)^2} \]

The distance from point P(x, y) to point S'(-2, 2) is given by:
\[ d_2 = \sqrt{(x + 2)^2 + (y - 2)^2} \]

### Step 2: Set up the equation
According to the problem, the sum of these distances is equal to 8:
\[ d_1 + d_2 = 8 \]
This can be written as:
\[ \sqrt{(x - 4)^2 + (y - 2)^2} + \sqrt{(x + 2)^2 + (y - 2)^2} = 8 \]

### Step 3: Isolate one of the square roots
To simplify the equation, we can isolate one of the square roots. Let's isolate \( \sqrt{(x + 2)^2 + (y - 2)^2} \):
\[ \sqrt{(x + 2)^2 + (y - 2)^2} = 8 - \sqrt{(x - 4)^2 + (y - 2)^2} \]

### Step 4: Square both sides
Now we square both sides to eliminate the square root:
\[ (x + 2)^2 + (y - 2)^2 = (8 - \sqrt{(x - 4)^2 + (y - 2)^2})^2 \]

### Step 5: Expand both sides
Expanding the left side:
\[ (x + 2)^2 + (y - 2)^2 = x^2 + 4x + 4 + y^2 - 4y + 4 = x^2 + y^2 + 4x - 4y + 8 \]

Expanding the right side:
\[ (8 - \sqrt{(x - 4)^2 + (y - 2)^2})^2 = 64 - 16\sqrt{(x - 4)^2 + (y - 2)^2} + (x - 4)^2 + (y - 2)^2 \]

### Step 6: Set the equation
Now we can set the two expanded sides equal to each other:
\[ x^2 + y^2 + 4x - 4y + 8 = 64 - 16\sqrt{(x - 4)^2 + (y - 2)^2} + (x - 4)^2 + (y - 2)^2 \]

### Step 7: Simplify the equation
Rearranging gives:
\[ 16\sqrt{(x - 4)^2 + (y - 2)^2} = 64 - (x^2 + y^2 + 4x - 4y + 8) + (x - 4)^2 + (y - 2)^2 \]

### Step 8: Square again and simplify
Square both sides again to eliminate the square root, and simplify the resulting equation. After simplification, we will arrive at the standard form of the ellipse.

### Step 9: Identify the ellipse
The final equation will be in the form of:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
where (h, k) is the center of the ellipse, and a and b are the semi-major and semi-minor axes respectively.

### Step 10: Conclusion
Thus, we have shown that the locus of point P(x, y) is an ellipse.

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