Let the focus S of the parabola `y^2=8x` lies on the focal chord PQ of the same parabola . If PS = 6 , then the square of the slope of the chord PQ is

Options

A. `2/sqrt5`
B. `4/5`
C. `5/4`
D. `9/4`

Answer

To solve the problem, we need to find the square of the slope of the focal chord PQ of the parabola given by the equation \( y^2 = 8x \). Let's go through the solution step by step.

### Step 1: Identify the parameters of the parabola
The given parabola is \( y^2 = 8x \). This can be compared to the standard form \( y^2 = 4ax \), where \( 4a = 8 \). Thus, we find:
\[ a = 2 \]

### Step 2: Determine the focus of the parabola
The focus \( S \) of the parabola \( y^2 = 8x \) is located at the point \( (a, 0) \). Therefore, the coordinates of the focus \( S \) are:
\[ S = (2, 0) \]

### Step 3: Parameterize a point \( P \) on the parabola
A point \( P \) on the parabola can be represented using the parameter \( t \). The coordinates of point \( P \) are given by:
\[ P = (2t^2, 4t) \]

### Step 4: Use the distance formula to find the relationship between \( P \) and \( S \)
We know that the distance \( PS \) is given as 6. Using the distance formula:
\[ PS = \sqrt{(2t^2 - 2)^2 + (4t - 0)^2} \]
Setting this equal to 6:
\[ \sqrt{(2t^2 - 2)^2 + (4t)^2} = 6 \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (2t^2 - 2)^2 + (4t)^2 = 36 \]
Expanding both terms:
\[ (2t^2 - 2)^2 = 4(t^2 - 1)^2 = 4(t^4 - 2t^2 + 1) \]
\[ (4t)^2 = 16t^2 \]
Combining these:
\[ 4(t^4 - 2t^2 + 1) + 16t^2 = 36 \]
Simplifying:
\[ 4t^4 + 8t^2 + 4 - 36 = 0 \]
\[ 4t^4 + 8t^2 - 32 = 0 \]

### Step 6: Divide by 4 to simplify
\[ t^4 + 2t^2 - 8 = 0 \]
Let \( u = t^2 \):
\[ u^2 + 2u - 8 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2} \]
This gives:
\[ u = 2 \quad \text{or} \quad u = -4 \]
Since \( u = t^2 \) must be non-negative, we have:
\[ t^2 = 2 \]

### Step 8: Find the coordinates of point \( P \)
Using \( t = \sqrt{2} \):
\[ P = (2t^2, 4t) = (2 \cdot 2, 4 \cdot \sqrt{2}) = (4, 4\sqrt{2}) \]

### Step 9: Calculate the slope of the chord \( PQ \)
The slope \( m \) of the chord \( PQ \) (which passes through \( P \) and \( S \)) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4\sqrt{2} - 0}{4 - 2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \]

### Step 10: Find the square of the slope
The square of the slope is:
\[ m^2 = (2\sqrt{2})^2 = 4 \cdot 2 = 8 \]

### Final Answer
The square of the slope of the chord \( PQ \) is:
\[ \boxed{8} \]