Let P be the point `(1,0)` and Q a point on the locus `y^2 = 8x`. The locus of mid-point of PQ is :

Options

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

Answer

To find the locus of the midpoint of the points P and Q, where P is the point (1, 0) and Q lies on the parabola defined by the equation \( y^2 = 8x \), we will follow these steps:

### Step 1: Define the Points
Let the coordinates of point Q on the parabola be \( (m, n) \). Since Q lies on the parabola \( y^2 = 8x \), it must satisfy this equation.

### Step 2: Substitute Q into the Parabola Equation
From the parabola equation, we have:
\[ n^2 = 8m \]
This implies that \( m = \frac{n^2}{8} \).

### Step 3: Find the Midpoint of P and Q
The coordinates of the midpoint A of the line segment PQ can be calculated using the midpoint formula:
\[ A = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of P (1, 0) and Q (m, n):
\[ H = \frac{1 + m}{2} \quad \text{and} \quad K = \frac{0 + n}{2} \]
Thus, the coordinates of the midpoint A are:
\[ H = \frac{1 + m}{2}, \quad K = \frac{n}{2} \]

### Step 4: Substitute m in Terms of n
Now, substitute \( m = \frac{n^2}{8} \) into the equation for H:
\[ H = \frac{1 + \frac{n^2}{8}}{2} = \frac{1}{2} + \frac{n^2}{16} \]

### Step 5: Express H in Terms of K
Since \( K = \frac{n}{2} \), we can express \( n \) in terms of \( K \):
\[ n = 2K \]
Substituting this into the equation for H:
\[ H = \frac{1}{2} + \frac{(2K)^2}{16} = \frac{1}{2} + \frac{4K^2}{16} = \frac{1}{2} + \frac{K^2}{4} \]

### Step 6: Rearrange the Equation
Now, we can rearrange the equation:
\[ H - \frac{1}{2} = \frac{K^2}{4} \]
Multiplying through by 4 gives:
\[ 4H - 2 = K^2 \]
Thus, we can express this as:
\[ K^2 = 4H - 2 \]

### Step 7: Write the Locus Equation
We can replace H and K with x and y respectively (where \( H = x \) and \( K = y \)):
\[ y^2 = 4x - 2 \]
This is the equation of the locus of the midpoint of PQ.

### Final Step: Rearranging the Locus Equation
We can rearrange this to the standard form:
\[ y^2 - 4x + 2 = 0 \]

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