Two perpendicular chords are drawn from the origin O to the parabola `y=x^2-x`, which meet the parabola at P and Q, then rectangle POQR is completed.
The locus of R is

Options

A. `y=2x^2+3x+4`
B. `y=x^2-3x+4`
C. `2y^2+3y+4=0`
D. `3y^2-3y+4=x`

Answer

To find the locus of point R in the given problem, we will follow these steps:

### Step 1: Understand the Parabola
The equation of the parabola is given as:
\[ y = x^2 - x \]
This can be rewritten as:
\[ y = x(x - 1) \]
The parabola opens upwards and intersects the x-axis at points (0,0) and (1,0).

### Step 2: Define Points P and Q
Let the coordinates of point P be \( P(\alpha, \alpha^2 - \alpha) \) and the coordinates of point Q be \( Q(\beta, \beta^2 - \beta) \). Here, \( \alpha \) and \( \beta \) are the x-coordinates where the chords intersect the parabola.

### Step 3: Slopes of Chords OP and OQ
The slope of line OP (from origin O to point P) is calculated as:
\[ m_{OP} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\alpha^2 - \alpha - 0}{\alpha - 0} = \frac{\alpha^2 - \alpha}{\alpha} = \alpha - 1 \]

Similarly, the slope of line OQ (from origin O to point Q) is:
\[ m_{OQ} = \frac{\beta^2 - \beta - 0}{\beta - 0} = \frac{\beta^2 - \beta}{\beta} = \beta - 1 \]

### Step 4: Condition for Perpendicularity
Since the chords OP and OQ are perpendicular, we have:
\[ m_{OP} \cdot m_{OQ} = -1 \]
Substituting the slopes:
\[ (\alpha - 1)(\beta - 1) = -1 \]
Expanding this gives:
\[ \alpha\beta - \alpha - \beta + 1 = -1 \]
Thus, we can simplify to:
\[ \alpha\beta - \alpha - \beta + 2 = 0 \]
Rearranging gives:
\[ \alpha\beta = \alpha + \beta - 2 \]
This is our first equation.

### Step 5: Midpoint of Diagonal PR
The coordinates of point R are \( R(H, K) \). The midpoint of diagonal PQ can be calculated as:
\[ M_{PQ} = \left( \frac{\alpha + \beta}{2}, \frac{(\alpha^2 - \alpha) + (\beta^2 - \beta)}{2} \right) \]

### Step 6: Midpoint of Diagonal OR
The midpoint of diagonal OR is:
\[ M_{OR} = \left( \frac{H + 0}{2}, \frac{K + 0}{2} \right) = \left( \frac{H}{2}, \frac{K}{2} \right) \]

### Step 7: Equating Midpoints
Since the midpoints of the diagonals of a rectangle are equal, we have:
\[ \frac{H}{2} = \frac{\alpha + \beta}{2} \]
This implies:
\[ H = \alpha + \beta \]

For the y-coordinates:
\[ \frac{K}{2} = \frac{(\alpha^2 - \alpha) + (\beta^2 - \beta)}{2} \]
This simplifies to:
\[ K = \alpha^2 - \alpha + \beta^2 - \beta \]

### Step 8: Express K in terms of H
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \), we can substitute:
\[ K = (H^2 - 2\alpha\beta) - H \]

### Step 9: Substitute \(\alpha\beta\)
From our earlier equation, substitute \(\alpha\beta\):
\[ K = H^2 - 2(H - 2) - H \]
This simplifies to:
\[ K = H^2 - H + 4 \]

### Step 10: Final Locus Equation
Thus, we have the equation of the locus:
\[ K = H^2 - H + 4 \]
Replacing \( K \) with \( y \) and \( H \) with \( x \):
\[ y = x^2 - x + 4 \]

### Conclusion
The locus of point R is given by:
\[ y = x^2 - x + 4 \]