If P is the point (1,0) and Q is any point on the parabola `y^(2) = 8x` then the locus of mid - point of PQ is

Options

A. `y^(2) - 4x + 2 = 0 `
B. ` x^(2) - 4y + 2 = 0 `
C. ` y^(2) - 4x - 2 = 0 `
D. ` x^(2) - 4y - 2 = 0 `

Answer

To find the locus of the midpoint of the line segment PQ, where P is the point (1, 0) and Q is any point on the parabola \( y^2 = 8x \), we can follow these steps:

### Step 1: Define the coordinates of points P and Q
- Let \( P = (1, 0) \)
- Let \( Q = (x', y') \) where \( Q \) lies on the parabola \( y^2 = 8x \).

### Step 2: Write the midpoint formula
The coordinates of the midpoint \( M \) of the segment \( PQ \) can be calculated using the midpoint formula:
\[ M = \left( \frac{x + x'}{2}, \frac{y + y'}{2} \right) \]
Substituting the coordinates of points P and Q:
\[ M = \left( \frac{1 + x'}{2}, \frac{0 + y'}{2} \right) = \left( \frac{1 + x'}{2}, \frac{y'}{2} \right) \]

### Step 3: Express \( x' \) and \( y' \) in terms of \( H \) and \( K \)
Let \( H = \frac{1 + x'}{2} \) and \( K = \frac{y'}{2} \).
From these, we can express \( x' \) and \( y' \):
\[ x' = 2H - 1 \]
\[ y' = 2K \]

### Step 4: Substitute \( x' \) and \( y' \) into the parabola equation
Since \( Q \) lies on the parabola \( y^2 = 8x \), we substitute \( x' \) and \( y' \) into this equation:
\[ (2K)^2 = 8(2H - 1) \]
This simplifies to:
\[ 4K^2 = 16H - 8 \]

### Step 5: Rearrange the equation
Dividing the entire equation by 4 gives:
\[ K^2 = 4H - 2 \]

### Step 6: Write the final equation
Now, substituting back \( K \) and \( H \) with \( y \) and \( x \) respectively, we have:
\[ y^2 = 4x - 2 \]
Rearranging gives us the final form of the equation:
\[ y^2 - 4x + 2 = 0 \]

### Conclusion
Thus, the locus of the midpoint of PQ is given by the equation:
\[ y^2 - 4x + 2 = 0 \]