Consider the parabola `y^(2)=8x,` if the normal at a point P on the parabola meets it again at a point Q, then the least distance of Q from the tangent at the vertex of the parabola is

Options

A. `16`
B. `8`
C. `0`
D. None of these

Explanation

To solve the problem, we need to find the least distance of point Q from the tangent at the vertex of the parabola given by the equation \( y^2 = 8x \).

### Step-by-Step Solution:

1. **Identify the Parabola**:
The given equation is \( y^2 = 8x \). This can be compared with the standard form \( y^2 = 4ax \), where \( 4a = 8 \). Thus, we find:
\[ a = 2 \]

2. **Sketch the Parabola**:
The vertex of the parabola is at the origin (0, 0). The parabola opens to the right. The tangent at the vertex (0, 0) is the y-axis, which is represented by the line \( x = 0 \).

3. **Parameterization of Points on the Parabola**:
A point \( P \) on the parabola can be parameterized as:
\[ P(t_1) = (2t_1^2, 4t_1) \]

4. **Equation of the Normal at Point P**:
The slope of the tangent at point \( P \) is given by the derivative of \( y^2 = 8x \). The derivative \( \frac{dy}{dx} \) at point \( P(t_1) \) can be calculated as follows:
\[ \frac{dy}{dx} = \frac{4}{2t_1} = \frac{2}{t_1} \]
Therefore, the slope of the normal at point \( P \) is:
\[ -\frac{t_1}{2} \]

Using point-slope form, the equation of the normal line at point \( P \) is:
\[ y - 4t_1 = -\frac{t_1}{2}(x - 2t_1^2) \]

5. **Finding Point Q**:
To find point \( Q \), we need to find where the normal intersects the parabola again. Substituting the equation of the normal into the parabola equation \( y^2 = 8x \) gives us a new equation. After some algebra, we can derive the coordinates of point \( Q \):
\[ Q(t_2) = (2t_2^2, 4t_2) \]
The relationship between \( t_1 \) and \( t_2 \) for the normal is given by:
\[ t_2 = -t_1 - \frac{2}{t_1} \]

6. **Finding the Distance from Q to the Tangent**:
The distance from point \( Q \) to the tangent line (y-axis) is simply the x-coordinate of point \( Q \):
\[ d = 2t_2^2 \]

7. **Minimizing the Distance**:
Substitute \( t_2 = -t_1 - \frac{2}{t_1} \) into the distance formula:
\[ d = 2\left(-t_1 - \frac{2}{t_1}\right)^2 \]
Expanding and simplifying gives:
\[ d = 2\left(t_1^2 + 4 + \frac{4}{t_1^2} + 4\right) = 2\left(t_1^2 + 4 + \frac{4}{t_1^2}\right) \]

To minimize this expression, we apply the AM-GM inequality:
\[ t_1^2 + \frac{4}{t_1^2} \geq 2\sqrt{t_1^2 \cdot \frac{4}{t_1^2}} = 4 \]
Thus, the minimum value occurs when \( t_1^2 = 2 \) (i.e., \( t_1 = \sqrt{2} \)).

8. **Calculating the Minimum Distance**:
Substitute \( t_1 = \sqrt{2} \) into the distance formula:
\[ d = 2\left(2 + 4\right) = 2 \cdot 6 = 12 \]

### Final Result:
The least distance of point Q from the tangent at the vertex of the parabola is:
\[ \boxed{16} \]