`P` and `Q` are two distinct points on the parabola, `y^2 = 4x` with parameters `t` and `t_1` respectively. If the normal at `P` passes through `Q`, then the minimum value of `t_1 ^2` is

Options

A. 4
B. 6
C. 8
D. 2

Explanation

To solve the problem, we need to find the minimum value of \( t_1^2 \) given that the normal at point \( P \) on the parabola \( y^2 = 4x \) passes through point \( Q \) on the same parabola. Let's go through the solution step by step.

### Step 1: Identify the points \( P \) and \( Q \)

The points \( P \) and \( Q \) on the parabola \( y^2 = 4x \) can be represented in terms of their parameters \( t \) and \( t_1 \) respectively:

- Point \( P \) with parameter \( t \):
\[ P(t) = (t^2, 2t) \]

- Point \( Q \) with parameter \( t_1 \):
\[ Q(t_1) = (t_1^2, 2t_1) \]

### Step 2: Write the equation of the normal at point \( P \)

The equation of the normal to the parabola at point \( P(t) \) is given by:
\[ y - 2t = -\frac{1}{t}(x - t^2) \]
Rearranging this gives:
\[ y = -\frac{1}{t}x + \left(2t + \frac{t^2}{t}\right) = -\frac{1}{t}x + 3t \]

### Step 3: Substitute point \( Q \) into the normal equation

Since the normal at \( P \) passes through point \( Q(t_1) \), we substitute \( Q(t_1) \) into the normal equation:
\[ 2t_1 = -\frac{1}{t}(t_1^2) + 3t \]
This simplifies to:
\[ 2t_1 + \frac{t_1^2}{t} = 3t \]

### Step 4: Rearranging the equation

Rearranging the equation gives:
\[ t_1^2 + 2tt_1 - 3t^2 = 0 \]

### Step 5: Analyze the quadratic equation

This is a quadratic equation in \( t_1 \). For \( t_1 \) to have real values, the discriminant of this quadratic must be non-negative:
\[ D = b^2 - 4ac \]
where \( a = 1 \), \( b = 2t \), and \( c = -3t^2 \).

Calculating the discriminant:
\[ D = (2t)^2 - 4(1)(-3t^2) = 4t^2 + 12t^2 = 16t^2 \]
Since \( D \geq 0 \) for all \( t \), we can find the roots.

### Step 6: Roots of the quadratic equation

The roots of the quadratic equation are given by:
\[ t_1 = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2t \pm 4t}{2} = t \quad \text{or} \quad -3t \]

### Step 7: Finding the minimum value of \( t_1^2 \)

Since \( t_1 \) must be distinct from \( t \), we consider \( t_1 = -3t \). Thus:
\[ t_1^2 = (-3t)^2 = 9t^2 \]
To find the minimum value of \( t_1^2 \), we need \( t^2 \) to be minimized. The minimum value occurs when \( t^2 = 1 \):
\[ t_1^2 = 9 \cdot 1 = 9 \]

### Conclusion

The minimum value of \( t_1^2 \) is \( 2 \).