If the sum of squares of distances of the point (x, y) from the points (a, 0) and (-a, 0) is 2b², then :

Question

Accepted Answer

To solve this problem, we need to analyze the condition given: the sum of the squares of the distances from the point $(x, y)$ to the points $(a, 0)$ and $(-a, 0)$ is $2b^2$.

1. **Calculate the distance from $(x, y)$ to $(a, 0)$:**

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
   $$
   \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
   $$

So, the distance from $(x, y)$ to $(a, 0)$ is:
   $$
   \sqrt{(x - a)^2 + y^2}
   $$

The square of this distance is:
   $$
   (x - a)^2 + y^2
   $$

2. **Calculate the distance from $(x, y)$ to $(-a, 0)$:**

Similarly, the distance from $(x, y)$ to $(-a, 0)$ is:
   $$
   \sqrt{(x + a)^2 + y^2}
   $$

The square of this distance is:
   $$
   (x + a)^2 + y^2
   $$

3. **Sum of the squares of the distances:**

According to the problem, the sum of these squares is $2b^2$:
   $$
   (x - a)^2 + y^2 + (x + a)^2 + y^2 = 2b^2
   $$

Simplifying the left side:
   $$
   (x^2 - 2ax + a^2) + y^2 + (x^2 + 2ax + a^2) + y^2 = 2b^2
   $$

Combine like terms:
   $$
   2x^2 + 2a^2 + 2y^2 = 2b^2
   $$

Divide the entire equation by 2:
   $$
   x^2 + a^2 + y^2 = b^2
   $$

Rearrange to match the form given in Option 4:
   $$
   x^2 + a^2 = b^2 - y^2
   $$

This matches exactly with Option 4: $x^2 + a^2 = b^2 - y^2$.

**Explanation for Incorrect Options:**

- **Option 1: $x^2 + a^2 = b^2 + y^2$**

This option suggests that the sum of $x^2 + a^2$ is equal to $b^2 + y^2$, which contradicts the derived equation $x^2 + a^2 = b^2 - y^2$.

- **Option 2: $x^2 + a^2 = 26^2 - y^2$**

This option introduces an arbitrary constant $26^2$ instead of $b^2$, which is not supported by the given condition $2b^2$.

- **Option 3: $x^2 - a^2 = b^2 + y^2$**

This option incorrectly suggests subtracting $a^2$ from $x^2$, which does not align with the derived equation $x^2 + a^2 = b^2 - y^2$.

Thus, Option 4 is the correct answer as it accurately reflects the relationship derived from the given condition.