Find the equation to the locus of a point P whose distance to (2,0)is equal to its distance from y-axis.

Answer

To find the equation of the locus of a point \( P(x, y) \) whose distance to the point \( (2, 0) \) is equal to its distance from the y-axis, we can follow these steps:

### Step 1: Set up the distances
The distance from point \( P(x, y) \) to the point \( (2, 0) \) can be calculated using the distance formula:
\[ d_1 = \sqrt{(x - 2)^2 + (y - 0)^2} = \sqrt{(x - 2)^2 + y^2} \]

The distance from point \( P(x, y) \) to the y-axis is simply the absolute value of the x-coordinate:
\[ d_2 = |x| \]

### Step 2: Set the distances equal
According to the problem, these two distances are equal:
\[ \sqrt{(x - 2)^2 + y^2} = |x| \]

### Step 3: Square both sides
To eliminate the square root, we square both sides of the equation:
\[ (x - 2)^2 + y^2 = x^2 \]

### Step 4: Expand the left side
Now, expand the left side of the equation:
\[ (x^2 - 4x + 4) + y^2 = x^2 \]

### Step 5: Simplify the equation
Subtract \( x^2 \) from both sides:
\[ -4x + 4 + y^2 = 0 \]

Rearranging gives us:
\[ y^2 - 4x + 4 = 0 \]

### Final Equation
Thus, the equation of the locus is:
\[ y^2 - 4x + 4 = 0 \]

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