Through the vertex O of a parabola `y^2 = 4x` chords OP and OQ are drawn at right angles to one-another. Then for all positions of P, PQ cuts the axis of the parabola at a fixed point and the locus of the middle point of PQ is

Options

A. `y^2 = 2 (x-2)`
B. `y^2 = 2 (x-4)`
C. `y^2 = 2 (x-6)`
D. none

Answer

To solve the problem, we need to find the locus of the midpoint of the chord PQ of the parabola \( y^2 = 4x \) when OP and OQ are drawn at right angles to one another through the vertex O.

### Step-by-Step Solution:

1. **Identify the Parabola and Vertex**:
The given parabola is \( y^2 = 4x \). The vertex O of this parabola is at the origin, \( O(0, 0) \).

**Hint**: Remember that the vertex of a parabola in the form \( y^2 = 4ax \) is at the point \( (0, 0) \).

2. **Parameterization of Points P and Q**:
We can represent points P and Q on the parabola using parameters \( t_1 \) and \( t_2 \):
- Point \( P(t_1) \) has coordinates \( (t_1^2, 2t_1) \)
- Point \( Q(t_2) \) has coordinates \( (t_2^2, 2t_2) \)

**Hint**: Use the parameterization of the parabola to express points in terms of a single variable.

3. **Find Slopes of OP and OQ**:
The slope of line OP is given by:
\[ m_1 = \frac{2t_1 - 0}{t_1^2 - 0} = \frac{2}{t_1} \]
The slope of line OQ is given by:
\[ m_2 = \frac{2t_2 - 0}{t_2^2 - 0} = \frac{2}{t_2} \]

**Hint**: The slope of a line through two points can be calculated using the formula \( \frac{y_2 - y_1}{x_2 - x_1} \).

4. **Condition for Perpendicularity**:
Since OP and OQ are perpendicular, we have:
\[ m_1 \cdot m_2 = -1 \implies \left(\frac{2}{t_1}\right) \cdot \left(\frac{2}{t_2}\right) = -1 \]
This simplifies to:
\[ t_1 t_2 = -4 \]

**Hint**: For two lines to be perpendicular, the product of their slopes must equal -1.

5. **Midpoint of Chord PQ**:
The midpoint M of the chord PQ has coordinates:
\[ M\left(\frac{t_1^2 + t_2^2}{2}, \frac{2t_1 + 2t_2}{2}\right) = \left(\frac{t_1^2 + t_2^2}{2}, t_1 + t_2\right) \]
Let \( h = \frac{t_1^2 + t_2^2}{2} \) and \( k = t_1 + t_2 \).

**Hint**: The midpoint formula for a line segment between two points can be used here.

6. **Express \( t_1^2 + t_2^2 \)**:
Using the identity \( (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2 \), we can express \( t_1^2 + t_2^2 \):
\[ t_1^2 + t_2^2 = k^2 - 2(-4) = k^2 + 8 \]

**Hint**: Remember the algebraic identity for the square of a sum.

7. **Substituting in Midpoint Coordinates**:
Substitute \( t_1^2 + t_2^2 \) into the equation for \( h \):
\[ h = \frac{k^2 + 8}{2} \]
Rearranging gives:
\[ k^2 = 2h - 8 \]

**Hint**: Rearranging equations can help isolate variables.

8. **Final Locus Equation**:
The equation \( k^2 = 2h - 8 \) can be rewritten as:
\[ k^2 = 2h - 8 \implies k^2 = 2(h - 4) \]
This represents a parabola in the form \( y^2 = 2(x - 4) \).

**Hint**: Recognizing the standard form of a parabola can help in identifying the locus.

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