The tangent and normal at the point `P(4,4)` to the parabola, `y^(2) = 4x` intersect the x-axis at the points Q and R, respectively. Then the circumcentre of the `DeltaPQR` is (a) `(2,0)` (b) `(2,1)` (c) `(1,0)` (d) `(1,2)`

Options

A. `(2,0)`
B. `(2,1)`
C. `(1,0)`
D. `(1,2)`

Explanation

To find the circumcenter of triangle PQR formed by the points P(4,4), Q, and R, where Q and R are the points of intersection of the tangent and normal to the parabola \(y^2 = 4x\) with the x-axis, we can follow these steps:

### Step 1: Identify the parabola and point P
The given parabola is \(y^2 = 4x\). The point \(P\) is given as \(P(4, 4)\).

### Step 2: Find the value of \(a\)
The standard form of the parabola is \(y^2 = 4ax\). Comparing this with the given equation, we find:
- \(4a = 4 \Rightarrow a = 1\).

### Step 3: Write the equation of the tangent at point P
The equation of the tangent to the parabola at point \((x_1, y_1)\) is given by:
\[ yy_1 = 2a(x + x_1) \]
Substituting \(x_1 = 4\), \(y_1 = 4\), and \(a = 1\):
\[ y \cdot 4 = 2 \cdot 1 \cdot (x + 4) \]
This simplifies to:
\[ 4y = 2x + 8 \quad \Rightarrow \quad 2x - 4y + 8 = 0 \]
or
\[ y = \frac{1}{2}x + 2 \]

### Step 4: Find the intersection of the tangent with the x-axis
To find the intersection point \(Q\), set \(y = 0\):
\[ 0 = \frac{1}{2}x + 2 \Rightarrow \frac{1}{2}x = -2 \Rightarrow x = -4 \]
Thus, the coordinates of point \(Q\) are \(Q(-4, 0)\).

### Step 5: Write the equation of the normal at point P
The slope of the tangent line \(m_1\) is \(\frac{1}{2}\). The slope of the normal \(m_2\) is given by:
\[ m_2 = -\frac{1}{m_1} = -2 \]
The equation of the normal at point \(P(4, 4)\) is:
\[ y - y_1 = m_2(x - x_1) \Rightarrow y - 4 = -2(x - 4) \]
This simplifies to:
\[ y - 4 = -2x + 8 \quad \Rightarrow \quad 2x + y - 12 = 0 \]

### Step 6: Find the intersection of the normal with the x-axis
To find the intersection point \(R\), set \(y = 0\):
\[ 2x + 0 - 12 = 0 \Rightarrow 2x = 12 \Rightarrow x = 6 \]
Thus, the coordinates of point \(R\) are \(R(6, 0)\).

### Step 7: Identify the points P, Q, and R
We have:
- \(P(4, 4)\)
- \(Q(-4, 0)\)
- \(R(6, 0)\)

### Step 8: Find the circumcenter of triangle PQR
The circumcenter is the midpoint of the line segment \(QR\) since \(Q\) and \(R\) lie on the x-axis and form the diameter of the circumcircle.

The midpoint \(S\) of segment \(QR\) is given by:
\[ S = \left(\frac{x_Q + x_R}{2}, \frac{y_Q + y_R}{2}\right) = \left(\frac{-4 + 6}{2}, \frac{0 + 0}{2}\right) = \left(\frac{2}{2}, 0\right) = (1, 0) \]

### Final Answer
Thus, the circumcenter of triangle \(PQR\) is \((1, 0)\).