The normal to the parabola `y^(2)=4x` at `P(9, 6)` meets the parabola again at Q. If the tangent at Q meets the directrix at R, then the slope of another tangent drawn from point R to this parabola is

Options

A. `11`
B. `(11)/(3)`
C. `(3)/(11)`
D. `3`

Answer

To solve the problem step-by-step, we will follow the given information and derive the necessary points and slopes.

### Step 1: Identify the parameters of the parabola
The equation of the parabola is given as \( y^2 = 4x \). Here, we can identify that \( a = 1 \) since the standard form is \( y^2 = 4ax \).

### Step 2: Find the parameter \( t_1 \) for point \( P(9, 6) \)
Using the parametric equations for the parabola, we have:
- \( x = at^2 \)
- \( y = 2at \)

For point \( P(9, 6) \):
- \( 9 = 1 \cdot t_1^2 \) implies \( t_1^2 = 9 \) so \( t_1 = 3 \).
- \( 6 = 2 \cdot 1 \cdot t_1 \) implies \( 2t_1 = 6 \) so \( t_1 = 3 \).

Thus, \( t_1 = 3 \).

### Step 3: Calculate the parameter \( t_2 \) for point \( Q \)
The normal at point \( P \) meets the parabola again at point \( Q \). The relationship between \( t_1 \) and \( t_2 \) for the normal is given by:
\[ t_2 = -t_1 - \frac{2}{t_1} \]
Substituting \( t_1 = 3 \):
\[ t_2 = -3 - \frac{2}{3} = -\frac{9}{3} - \frac{2}{3} = -\frac{11}{3} \]

### Step 4: Find the coordinates of point \( Q \)
Using \( t_2 = -\frac{11}{3} \):
- \( x_Q = a t_2^2 = 1 \cdot \left(-\frac{11}{3}\right)^2 = \frac{121}{9} \)
- \( y_Q = 2a t_2 = 2 \cdot 1 \cdot \left(-\frac{11}{3}\right) = -\frac{22}{3} \)

Thus, the coordinates of point \( Q \) are \( Q\left(\frac{121}{9}, -\frac{22}{3}\right) \).

### Step 5: Find the tangent at point \( Q \)
The equation of the tangent at point \( Q(t_2) \) is given by:
\[ t_2 y = x + at_2^2 \]
Substituting \( t_2 = -\frac{11}{3} \) and \( a = 1 \):
\[ -\frac{11}{3} y = x + 1 \cdot \left(-\frac{11}{3}\right)^2 \]
Calculating \( \left(-\frac{11}{3}\right)^2 = \frac{121}{9} \):
\[ -\frac{11}{3} y = x + \frac{121}{9} \]
Rearranging gives:
\[ x + \frac{11}{3} y = \frac{121}{9} \]

### Step 6: Find the directrix and point \( R \)
The directrix of the parabola \( y^2 = 4x \) is given by \( x = -1 \). To find point \( R \), we need to find where the tangent at \( Q \) intersects the directrix \( x = -1 \):
Substituting \( x = -1 \) into the tangent equation:
\[ -1 + \frac{11}{3} y = \frac{121}{9} \]
Solving for \( y \):
\[ \frac{11}{3} y = \frac{121}{9} + 1 = \frac{121}{9} + \frac{9}{9} = \frac{130}{9} \]
Thus,
\[ y = \frac{130}{9} \cdot \frac{3}{11} = \frac{390}{99} = \frac{130}{33} \]
So, point \( R \) is \( R(-1, \frac{130}{33}) \).

### Step 7: Find the slope of another tangent from point \( R \)
Let \( t_3 \) be the parameter for the tangent from point \( R \). The condition for the tangents from \( R \) is:
\[ t_2 t_3 = -1 \]
Substituting \( t_2 = -\frac{11}{3} \):
\[ -\frac{11}{3} t_3 = -1 \implies t_3 = \frac{3}{11} \]

### Step 8: Find the slope of the tangent at point \( S \)
The slope of the tangent at point \( S \) (where \( t = t_3 \)) is given by:
\[ \text{slope} = 2a = 2 \cdot 1 = 2 \]

Thus, the slope of the tangent drawn from point \( R \) to the parabola is \( \frac{11}{3} \).

### Final Answer
The slope of another tangent drawn from point \( R \) to the parabola is \( \frac{11}{3} \).