Let PQ be the latus rectum of the parabola `y^2 = 4x` with vetex A. Minimum length of the projection of PQ on a tangent drawn in portion of Parabola PAQ is

Options

A. `2sqrt(2)`
B. `2asqrt(2)`
C. `2`
D. `3asqrt(2)`

Explanation

To solve the problem, we need to find the minimum length of the projection of the latus rectum \( PQ \) of the parabola \( y^2 = 4x \) onto a tangent drawn at a point on the parabola. Here are the steps to arrive at the solution:

### Step 1: Identify the Latus Rectum Points
The latus rectum of the parabola \( y^2 = 4x \) is a line segment parallel to the y-axis that passes through the focus of the parabola. The focus of the parabola is at the point \( (1, 0) \). The endpoints of the latus rectum \( PQ \) can be found by substituting \( x = 1 \) into the parabola's equation:
\[ y^2 = 4(1) \implies y^2 = 4 \implies y = 2 \text{ or } y = -2. \]
Thus, the points \( P \) and \( Q \) are \( P(1, 2) \) and \( Q(1, -2) \).

### Step 2: Find the Equation of the Tangent
The equation of the tangent line to the parabola \( y^2 = 4x \) at a point \( (x_0, y_0) \) on the parabola is given by:
\[ yy_0 = 2(x + x_0). \]
We can choose a point on the parabola, say \( (1, 2) \) (point \( P \)), to find the tangent line at this point:
\[ y(2) = 2(x + 1) \implies 2y = 2x + 2 \implies y = x + 1. \]

### Step 3: Determine the Angle of the Tangent
The slope of the tangent line \( y = x + 1 \) is \( 1 \). The angle \( \theta \) that this line makes with the x-axis can be found using:
\[ \tan(\theta) = \text{slope} = 1 \implies \theta = 45^\circ. \]

### Step 4: Calculate the Length of the Projection
The length of the latus rectum \( PQ \) is the distance between points \( P(1, 2) \) and \( Q(1, -2) \):
\[ \text{Length of } PQ = |2 - (-2)| = 4. \]
The projection of line segment \( PQ \) onto the tangent line can be calculated using the formula:
\[ \text{Projection length} = \text{Length of } PQ \cdot \cos(\theta). \]
Substituting the known values:
\[ \text{Projection length} = 4 \cdot \cos(45^\circ) = 4 \cdot \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}. \]

### Final Answer
Thus, the minimum length of the projection of \( PQ \) on the tangent drawn at point \( P \) is:
\[ \boxed{2\sqrt{2}}. \]