Let a, r, s, t be non-zero real numbers. Let `P(at^2, 2at), Q, R(ar^2, 2ar) and S(as^2, 2as)` be distinct points onthe parabola `y^2 = 4ax`. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K isthe point (2a, 0). The value of r is

Options

A. `-(1)/(t)`
B. `(t^(2)+1)/(t)`
C. `(1)/(t)`
D. `(t^(2)-1)/(t)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the coordinates of the points on the parabola
Given the parabola \( y^2 = 4ax \), the points are defined as follows:
- \( P(at^2, 2at) \)
- \( Q \) (unknown coordinates)
- \( R(ar^2, 2ar) \)
- \( S(as^2, 2as) \)

### Step 2: Use the property of focal chords
For points \( P \) and \( Q \) to be endpoints of a focal chord, we know that if \( t_1 \) and \( t_2 \) are the parameters of the endpoints of a focal chord, then:
\[ t_1 \cdot t_2 = -1 \]
Here, let \( t_1 = t \) for point \( P \) and \( t_2 = -\frac{1}{t} \) for point \( Q \).

### Step 3: Determine the coordinates of point \( Q \)
Using the parameter \( t_2 = -\frac{1}{t} \):
\[ Q\left(a\left(-\frac{1}{t}\right)^2, 2a\left(-\frac{1}{t}\right)\right) = \left(\frac{a}{t^2}, -\frac{2a}{t}\right) \]

### Step 4: Identify the coordinates of point \( K \)
Point \( K \) is given as \( (2a, 0) \).

### Step 5: Calculate the slopes of the lines
The slope of the line \( QR \) can be calculated using the coordinates of points \( Q \) and \( R \):
\[ \text{slope of } QR = \frac{2ar - \left(-\frac{2a}{t}\right)}{ar^2 - \frac{a}{t^2}} = \frac{2ar + \frac{2a}{t}}{ar^2 - \frac{a}{t^2}} = \frac{2a\left(r + \frac{1}{t}\right)}{a\left(r^2 - \frac{1}{t^2}\right)} = \frac{2\left(r + \frac{1}{t}\right)}{r^2 - \frac{1}{t^2}} \]

The slope of the line \( PK \) is:
\[ \text{slope of } PK = \frac{2at - 0}{at^2 - 2a} = \frac{2t}{t^2 - 2} \]

### Step 6: Set the slopes equal
Since lines \( QR \) and \( PK \) are parallel, their slopes are equal:
\[ \frac{2\left(r + \frac{1}{t}\right)}{r^2 - \frac{1}{t^2}} = \frac{2t}{t^2 - 2} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ 2\left(r + \frac{1}{t}\right)(t^2 - 2) = 2t\left(r^2 - \frac{1}{t^2}\right) \]
Dividing both sides by 2:
\[ \left(r + \frac{1}{t}\right)(t^2 - 2) = t\left(r^2 - \frac{1}{t^2}\right) \]

### Step 8: Rearranging and solving for \( r \)
Expanding both sides:
\[ rt^2 - 2r + \frac{t^2}{t} - 2\frac{1}{t} = tr^2 - \frac{1}{t} \]
Rearranging gives:
\[ rt^2 - tr^2 - 2r + \frac{1}{t} - 2\frac{1}{t} = 0 \]
This simplifies to:
\[ r(t^2 - t) = 2r - \frac{1}{t} \]
Solving for \( r \):
\[ r = \frac{t^2 - 1}{t} \]

### Final Answer
Thus, the value of \( r \) is:
\[ \boxed{\frac{t^2 - 1}{t}} \]