If the locus of the moving point P(x,y) satisfying `sqrt((x-1)^(2)+y^(2))+sqrt((x+1)^(2)+(y-sqrt(12))^(2))=a` is an ellipse, then find the values of a.

Explanation

To solve the problem, we need to analyze the given equation and determine the values of \( a \) for which the locus of the point \( P(x, y) \) forms an ellipse.

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
The equation given is:
\[ \sqrt{(x-1)^2 + y^2} + \sqrt{(x+1)^2 + (y - \sqrt{12})^2} = a \]
This represents the sum of the distances from the point \( P(x, y) \) to two fixed points \( S_1(1, 0) \) and \( S_2(-1, \sqrt{12}) \).

2. **Identifying the Foci**:
The points \( S_1 \) and \( S_2 \) are the foci of the ellipse:
- \( S_1(1, 0) \)
- \( S_2(-1, \sqrt{12}) \)

3. **Calculating the Distance Between the Foci**:
The distance \( d \) between the two foci \( S_1 \) and \( S_2 \) can be calculated using the distance formula:
\[ d = \sqrt{(1 - (-1))^2 + (0 - \sqrt{12})^2} \]
Simplifying this, we have:
\[ d = \sqrt{(1 + 1)^2 + (0 - \sqrt{12})^2} = \sqrt{2^2 + (-\sqrt{12})^2} = \sqrt{4 + 12} = \sqrt{16} = 4 \]

4. **Condition for the Locus to be an Ellipse**:
For the locus to be an ellipse, the sum of the distances from any point \( P \) to the two foci must be greater than the distance between the foci:
\[ a > d \]
Therefore, we need:
\[ a > 4 \]

5. **Conclusion**:
The values of \( a \) for which the locus of the point \( P(x, y) \) is an ellipse are:
\[ a > 4 \]
This means \( a \) can take any value greater than 4, such as 5, 6, 7, and so on.