The point on the line y = x such that the sum of the squares of its distance from the point (a, 0), (–a, 0) and (0, b) is minimum will be

Options

A. `(a//6 , a//6)`
B. `(a , a)`
C. `(b , b)`
D. ` (b //6 , b//6)`

Answer

To solve the problem of finding the point on the line \( y = x \) such that the sum of the squares of its distances from the points \( (a, 0) \), \( (-a, 0) \), and \( (0, b) \) is minimized, we can follow these steps:

### Step 1: Define the Point on the Line
Let the point on the line \( y = x \) be represented as \( P(t, t) \), where \( t \) is the x-coordinate (and also the y-coordinate since \( y = x \)).

### Step 2: Calculate the Distances
We need to calculate the distances from point \( P(t, t) \) to the three given points:
1. Distance to \( (a, 0) \):
\[ d_1 = \sqrt{(t - a)^2 + (t - 0)^2} = \sqrt{(t - a)^2 + t^2} \]
2. Distance to \( (-a, 0) \):
\[ d_2 = \sqrt{(t + a)^2 + (t - 0)^2} = \sqrt{(t + a)^2 + t^2} \]
3. Distance to \( (0, b) \):
\[ d_3 = \sqrt{(t - 0)^2 + (t - b)^2} = \sqrt{t^2 + (t - b)^2} \]

### Step 3: Square the Distances
Since we want the sum of the squares of these distances, we calculate:
1. \( d_1^2 = (t - a)^2 + t^2 = (t^2 - 2at + a^2 + t^2) = 2t^2 - 2at + a^2 \)
2. \( d_2^2 = (t + a)^2 + t^2 = (t^2 + 2at + a^2 + t^2) = 2t^2 + 2at + a^2 \)
3. \( d_3^2 = t^2 + (t - b)^2 = t^2 + (t^2 - 2bt + b^2) = 2t^2 - 2bt + b^2 \)

### Step 4: Formulate the Objective Function
Now, we sum these squared distances:
\[ S(t) = d_1^2 + d_2^2 + d_3^2 = (2t^2 - 2at + a^2) + (2t^2 + 2at + a^2) + (2t^2 - 2bt + b^2) \]
Combining like terms:
\[ S(t) = 6t^2 + 2a^2 + b^2 - 2bt \]

### Step 5: Differentiate and Find Critical Points
To minimize \( S(t) \), we take the derivative and set it to zero:
\[ S'(t) = 12t - 2b = 0 \]
Solving for \( t \):
\[ 12t = 2b \implies t = \frac{b}{6} \]

### Step 6: Verify Minimum
To confirm that this is a minimum, we take the second derivative:
\[ S''(t) = 12 \]
Since \( S''(t) > 0 \), this indicates a minimum at \( t = \frac{b}{6} \).

### Step 7: Determine the Point
The point \( P \) on the line \( y = x \) that minimizes the sum of the squares of the distances is:
\[ P\left(\frac{b}{6}, \frac{b}{6}\right) \]

### Final Answer
The point on the line \( y = x \) such that the sum of the squares of its distances from the points \( (a, 0) \), \( (-a, 0) \), and \( (0, b) \) is minimized is:
\[ \boxed{\left(\frac{b}{6}, \frac{b}{6}\right)} \]