A tangent is drawn to the parabola `y^(2)=8x` at P(2, 4) to intersect the x-axis at Q, from which another tangent is drawn to the parabola to touch it at R. If the normal at R intersects the parabola again at S, then the coordinates of S are

Options

A. (6, 4)
B. (18, 12)
C. (8, 8)
D. (8, 6)

Answer

To solve the problem step by step, we will follow the geometric properties of the parabola and the relationships between the points mentioned.

### Step 1: Identify the parabola and the point P
The given parabola is \( y^2 = 8x \). The point \( P(2, 4) \) lies on this parabola.

### Step 2: Find the equation of the tangent at point P
The formula for the equation of the tangent to the parabola \( y^2 = 4ax \) at point \( (x_1, y_1) \) is given by:
\[ yy_1 = 2a(x + x_1) \]
For our parabola \( y^2 = 8x \), we have \( a = 2 \). Thus, the equation of the tangent at point \( P(2, 4) \) is:
\[ y \cdot 4 = 2 \cdot 2 (x + 2) \]
Simplifying this gives:
\[ 4y = 8(x + 2) \implies 4y = 8x + 16 \implies y = 2x + 4 \]

### Step 3: Find the intersection of the tangent with the x-axis
To find point \( Q \), we set \( y = 0 \) in the tangent equation:
\[ 0 = 2x + 4 \implies 2x = -4 \implies x = -2 \]
Thus, the coordinates of point \( Q \) are \( (-2, 0) \).

### Step 4: Find the slope of the line QR
Next, we need to find the tangent line at point \( Q(-2, 0) \). The slope of the line \( PQ \) is:
\[ \text{slope} = \frac{0 - 4}{-2 - 2} = \frac{-4}{-4} = 1 \]
Since the slope of the tangent at \( Q \) will be the negative reciprocal of this slope, it will be \( -1 \).

### Step 5: Write the equation of the tangent at Q
Using the point-slope form of the line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( Q(-2, 0) \) and slope \( -1 \):
\[ y - 0 = -1(x + 2) \implies y = -x - 2 \]

### Step 6: Find the point R where this tangent touches the parabola
To find point \( R \), we set the equation of the tangent equal to the parabola:
\[ -x - 2 = \sqrt{8x} \quad \text{(taking the positive root)} \]
Squaring both sides:
\[ (-x - 2)^2 = 8x \implies x^2 + 4x + 4 = 8x \implies x^2 - 4x + 4 = 0 \]
Factoring gives:
\[ (x - 2)^2 = 0 \implies x = 2 \]
Substituting \( x = 2 \) back into the tangent equation to find \( y \):
\[ y = -2 - 2 = -4 \]
Thus, point \( R \) is \( (2, -4) \).

### Step 7: Find the normal at point R
The slope of the tangent at \( R(2, -4) \) is:
\[ \text{slope} = \frac{-4 - 0}{2 - (-2)} = \frac{-4}{4} = -1 \]
The slope of the normal is the negative reciprocal, which is \( 1 \). The equation of the normal at \( R \) is:
\[ y + 4 = 1(x - 2) \implies y = x - 6 \]

### Step 8: Find the intersection of the normal with the parabola again
Set the normal equation equal to the parabola:
\[ x - 6 = \sqrt{8x} \]
Squaring both sides:
\[ (x - 6)^2 = 8x \implies x^2 - 12x + 36 = 8x \implies x^2 - 20x + 36 = 0 \]
Using the quadratic formula:
\[ x = \frac{20 \pm \sqrt{400 - 144}}{2} = \frac{20 \pm \sqrt{256}}{2} = \frac{20 \pm 16}{2} \]
This gives:
\[ x = 18 \quad \text{or} \quad x = 2 \]
Since \( x = 2 \) corresponds to point \( R \), we take \( x = 18 \).

### Step 9: Find the corresponding y-coordinate for S
Substituting \( x = 18 \) back into the parabola:
\[ y^2 = 8 \cdot 18 = 144 \implies y = 12 \quad \text{or} \quad y = -12 \]
Thus, the coordinates of point \( S \) are \( (18, 12) \) and \( (18, -12) \).

### Final Answer
The coordinates of S are \( (18, 12) \) and \( (18, -12) \).