The normal to the parabola `y^(2)=4x` at P (1, 2) meets the parabola again in Q, then coordinates of Q are

Options

A. (-6, 9)
B. (9, -6)
C. (-9, -6)
D. (-6, -9)

Explanation

To find the coordinates of point Q where the normal to the parabola \( y^2 = 4x \) at point P(1, 2) meets the parabola again, we can follow these steps:

### Step 1: Identify the parabola and its parameters
The given parabola is \( y^2 = 4x \). This is a standard form of a parabola where \( a = 1 \).

### Step 2: Find the parameter \( t_1 \) for point P(1, 2)
For the parabola \( y^2 = 4ax \), the coordinates of a point on the parabola can be expressed in terms of the parameter \( t \) as:
\[ (x, y) = (at^2, 2at) \]
Substituting \( a = 1 \), we have:
\[ (x, y) = (t^2, 2t) \]
At point P(1, 2):
- From \( t^2 = 1 \), we get \( t_1 = 1 \) or \( t_1 = -1 \).
- From \( 2t = 2 \), we find \( t_1 = 1 \).

Thus, the parameter \( t_1 = 1 \).

### Step 3: Find the equation of the normal at point P
The equation of the normal to the parabola at the point corresponding to \( t_1 \) is given by:
\[ y - 2 = -\frac{1}{t_1}(x - 1) \]
Substituting \( t_1 = 1 \):
\[ y - 2 = -1(x - 1) \]
This simplifies to:
\[ y = -x + 3 \]

### Step 4: Find the intersection of the normal with the parabola
To find point Q, we need to solve the equations of the normal and the parabola simultaneously. Substitute \( y = -x + 3 \) into the parabola \( y^2 = 4x \):
\[ (-x + 3)^2 = 4x \]
Expanding this gives:
\[ x^2 - 6x + 9 = 4x \]
Rearranging it results in:
\[ x^2 - 10x + 9 = 0 \]

### Step 5: Solve the quadratic equation
We can solve the quadratic equation \( x^2 - 10x + 9 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -10, c = 9 \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} \]
\[ x = \frac{10 \pm \sqrt{100 - 36}}{2} \]
\[ x = \frac{10 \pm \sqrt{64}}{2} \]
\[ x = \frac{10 \pm 8}{2} \]
This gives us two solutions:
\[ x = \frac{18}{2} = 9 \quad \text{and} \quad x = \frac{2}{2} = 1 \]

### Step 6: Find corresponding y-coordinates
We already know that one intersection point is P(1, 2). Now, we find the y-coordinate for \( x = 9 \):
Substituting \( x = 9 \) into the normal equation:
\[ y = -9 + 3 = -6 \]
Thus, the coordinates of point Q are \( (9, -6) \).

### Final Answer
The coordinates of point Q are \( (9, -6) \).

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