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

Options

A. 1
B. `(1)/(2)`
C. `(3)/(4)`
D. `(3)/(2)`

Answer

To find the locus of the vertex R of the rectangle POQR formed by two perpendicular chords drawn from the origin O to the parabola \( y = x^2 \), we will follow these steps:

### Step 1: Define Points P and Q
Let the coordinates of point P be \( (a, a^2) \) and the coordinates of point Q be \( (b, b^2) \). Here, \( a \) and \( b \) are the x-coordinates where the chords intersect the parabola.

### Step 2: Find the Slopes of Chords
The slope of the line OP (from O to P) is given by:
\[ m_1 = \frac{a^2 - 0}{a - 0} = \frac{a^2}{a} = a \]
The slope of the line OQ (from O to Q) is given by:
\[ m_2 = \frac{b^2 - 0}{b - 0} = \frac{b^2}{b} = b \]

### Step 3: Use the Perpendicular Condition
Since the chords are perpendicular, the product of their slopes must equal -1:
\[ m_1 \cdot m_2 = -1 \implies a \cdot b = -1 \]

### Step 4: Find the Coordinates of R
The coordinates of point R, which is the vertex of the rectangle POQR, can be determined as follows:
- The x-coordinate of R is the average of the x-coordinates of P and Q:
\[ x_R = \frac{a + b}{2} \]
- The y-coordinate of R is the average of the y-coordinates of P and Q:
\[ y_R = \frac{a^2 + b^2}{2} \]

### Step 5: Express \( y_R \) in Terms of \( x_R \)
From the equation \( a \cdot b = -1 \), we can express \( b \) in terms of \( a \):
\[ b = -\frac{1}{a} \]
Now substituting \( b \) into the expressions for \( x_R \) and \( y_R \):
\[ x_R = \frac{a - \frac{1}{a}}{2} = \frac{a^2 - 1}{2a} \]
\[ y_R = \frac{a^2 + \left(-\frac{1}{a}\right)^2}{2} = \frac{a^2 + \frac{1}{a^2}}{2} \]

### Step 6: Substitute \( x_R \) into \( y_R \)
To find the relationship between \( y_R \) and \( x_R \), we can express \( a^2 \) in terms of \( x_R \):
\[ x_R = \frac{a^2 - 1}{2a} \implies 2a x_R = a^2 - 1 \implies a^2 = 2a x_R + 1 \]
Now substitute \( a^2 \) into the expression for \( y_R \):
\[ y_R = \frac{(2a x_R + 1) + \frac{1}{(2a x_R + 1)^2}}{2} \]
This expression can be simplified, but for the locus, we can derive a simpler form.

### Step 7: Final Locus Equation
Using \( a^2 + b^2 = (a + b)^2 - 2ab \):
\[ y_R = \frac{(a + b)^2 - 2(-1)}{2} = \frac{(a + b)^2 + 2}{2} \]
Substituting \( x_R \) gives us:
\[ y_R = \frac{x_R^2 + 2}{2} \]
Thus, the locus of point R is:
\[ y = \frac{x^2}{2} + 1 \]