If a normal subtends a right angle at the vertex of a parabola `y^(2)=4ax` then its length is

Options

A. `2sqrt(3)a`
B. `4sqrt(3)a`
C. `6sqrt(3)a`
D. `8sqrt(3)a`

Answer

To solve the problem of finding the length of the normal that subtends a right angle at the vertex of the parabola \( y^2 = 4ax \), we can follow these steps:

### Step 1: Understand the parabola and its properties
The given parabola is \( y^2 = 4ax \). The vertex of this parabola is at the origin (0, 0).

### Step 2: Parametric representation of points on the parabola
Any point \( P \) on the parabola can be represented in parametric form as:
\[ P(t_1) = (at_1^2, 2at_1) \]
Similarly, another point \( Q \) can be represented as:
\[ Q(t_2) = (at_2^2, 2at_2) \]

### Step 3: Equation of the normal to the parabola
The equation of the normal to the parabola at point \( P(t_1) \) is given by:
\[ y = -tx + 2at + at^2 \]
Substituting \( t = t_1 \), we have:
\[ y = -t_1x + 2at_1 + at_1^2 \]

### Step 4: Points \( P \) and \( Q \) lie on the normal
Since both points \( P \) and \( Q \) lie on the normal, they satisfy the normal's equation. Thus, we can substitute the coordinates of \( P \) and \( Q \) into the normal's equation.

### Step 5: Find the slopes of OP and OQ
The slopes of lines \( OP \) and \( OQ \) can be calculated as follows:
- Slope of \( OP \):
\[ \text{slope of } OP = \frac{2at_1}{at_1^2} = \frac{2}{t_1} \]
- Slope of \( OQ \):
\[ \text{slope of } OQ = \frac{2at_2}{at_2^2} = \frac{2}{t_2} \]

### Step 6: Use the condition of perpendicularity
Since the lines \( OP \) and \( OQ \) are perpendicular, we have:
\[ \left(\frac{2}{t_1}\right) \left(\frac{2}{t_2}\right) = -1 \]
This simplifies to:
\[ \frac{4}{t_1 t_2} = -1 \implies t_1 t_2 = -4 \]

### Step 7: Relate \( t_1 \) and \( t_2 \)
From the normal's equation, we can express \( t_2 \) in terms of \( t_1 \):
\[ t_2 = -\frac{t_1 + 2}{t_1} \]

### Step 8: Substitute and solve for \( t_1 \)
Substituting \( t_2 \) into the equation \( t_1 t_2 = -4 \):
\[ t_1 \left(-\frac{t_1 + 2}{t_1}\right) = -4 \implies -(t_1 + 2) = -4 \implies t_1 + 2 = 4 \implies t_1 = 2 \]

### Step 9: Find \( t_2 \)
Using \( t_1 = 2 \) in the relation for \( t_2 \):
\[ t_2 = -\frac{2 + 2}{2} = -2 \]

### Step 10: Calculate the length of the normal
The length \( PQ \) can be calculated using the distance formula:
\[ PQ = \sqrt{(2a(-2)^2 - 2a(2)^2)^2 + (2a(-2) - 2a(2))^2} \]
This simplifies to:
\[ = \sqrt{(8a - 8a)^2 + (-4a - 4a)^2} = \sqrt{0 + (-8a)^2} = 8a \]

### Final Answer
Thus, the length of the normal that subtends a right angle at the vertex of the parabola \( y^2 = 4ax \) is:
\[ \text{Length} = 6\sqrt{3}a \]