The sum of the squares of the distances of a moving point from two fixed points \((a, 0)\) and \((-a, 0)\) is equal to a constant quantity \(2c^2\). Then the equation of its locus is:

1. \(x^2 + y^2 = c^2 + a^2\)
2. \(x^2 + y^2 = c^2 - a^2\)
3. \(x^2 - y^2 = c^2 - a^2\)
4. [tex]\(x^2 - y^2 = c^2 + a^2\)[/tex]



Answer :

To determine the equation of the locus of a moving point whose sum of the squares of the distances from two fixed points is a constant, let's analyze the given condition step-by-step:

1. Define the points and variables:
Let the fixed points be \( (a, 0) \) and \( (-a, 0) \). Let \( (x, y) \) be the coordinates of the moving point.

2. Express the distances:
The distance from the moving point \( (x, y) \) to \( (a, 0) \) is given by:
[tex]\[ \sqrt{(x - a)^2 + y^2} \][/tex]
Similarly, the distance from the moving point \( (x, y) \) to \( (-a, 0) \) is:
[tex]\[ \sqrt{(x + a)^2 + y^2} \][/tex]

3. Apply the given condition:
According to the problem, the sum of the squares of these distances is \( 2c^2 \):
[tex]\[ \left( \sqrt{(x - a)^2 + y^2} \right)^2 + \left( \sqrt{(x + a)^2 + y^2} \right)^2 = 2c^2 \][/tex]

4. Simplify the equation:
[tex]\[ (x - a)^2 + y^2 + (x + a)^2 + y^2 = 2c^2 \][/tex]

5. Expand and combine like terms:
[tex]\[ x^2 - 2ax + a^2 + y^2 + x^2 + 2ax + a^2 + y^2 = 2c^2 \][/tex]

6. Combine the terms:
[tex]\[ x^2 + x^2 - 2ax + 2ax + a^2 + a^2 + y^2 + y^2 = 2c^2 \][/tex]
[tex]\[ 2x^2 + 2a^2 + 2y^2 = 2c^2 \][/tex]

7. Factor out the common term and divide by 2:
[tex]\[ x^2 + a^2 + y^2 = c^2 \][/tex]

8. Rearrange the equation:
[tex]\[ x^2 + y^2 = c^2 - a^2 \][/tex]

Therefore, the correct equation for the locus of the moving point under the given conditions is:
[tex]\[ x^2 + y^2 = c^2 - a^2 \][/tex]

Thus, the correct answer is:

2) [tex]\( x^2 + y^2 = c^2 - a^2 \)[/tex]