A normal drawn at a point P on the parabola `y^2 = 4ax` meets the curve again at Q. The least distance of Q from the axis of the parabola, is

Options

A. `2sqrt2a`
B. `3sqrt2a`
C. `4sqrta`
D. none of these

Explanation

To solve the problem, we need to find the least distance of point Q from the axis of the parabola given that a normal drawn at point P on the parabola \(y^2 = 4ax\) meets the curve again at Q.

### Step-by-Step Solution:

1. **Identify the point P on the parabola**:
Let the point P on the parabola be represented as \(P(at^2, 2at)\), where \(t\) is the parameter corresponding to point P.

**Hint**: The coordinates of any point on the parabola \(y^2 = 4ax\) can be expressed in parametric form.

2. **Equation of the normal at point P**:
The slope of the tangent at point P is given by \(\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}\). Therefore, the slope of the normal is \(-t\). The equation of the normal at point P is:
\[ y - 2at = -t(x - at^2) \]
Rearranging gives:
\[ y = -tx + at^2 + 2at \]

3. **Finding the intersection point Q**:
To find the point Q where the normal meets the parabola again, substitute the equation of the normal into the parabola's equation \(y^2 = 4ax\):
\[ (-tx + at^2 + 2at)^2 = 4ax \]
Expanding and simplifying will lead to a quadratic equation in \(x\).

4. **Using the property of normals**:
The condition for the normal at point P meeting the parabola again at point Q is given by:
\[ t_1 = -\frac{t + 2}{t} \]
where \(t_1\) is the parameter for point Q.

**Hint**: This condition arises from the geometric properties of the parabola and the normals.

5. **Finding the distance of Q from the axis**:
The distance of point Q from the axis of the parabola (the y-axis) is given by the x-coordinate of Q, which can be expressed in terms of \(t\):
\[ x_Q = a\left(-\frac{t + 2}{t}\right)^2 \]
Simplifying gives:
\[ x_Q = a\left(\frac{(t + 2)^2}{t^2}\right) = \frac{a(t^2 + 4t + 4)}{t^2} \]

6. **Minimizing the distance**:
To find the least distance, we need to minimize the expression for \(x_Q\). Using the AM-GM inequality:
\[ \frac{2a(t + 2)}{t} \geq 2a\sqrt{2} \]
This leads to the conclusion that the least distance from the axis of the parabola is:
\[ 4a\sqrt{2} \]

### Final Answer:
The least distance of Q from the axis of the parabola is \(4a\sqrt{2}\).

---