If a chord which is normal to the parabola at one end subtend a right angle at the vertex, then angle to the axis is (a) `alpha=sqrt(2)` (b) `alpha=sqrt(3)` (c) `alpha=(1)/(sqrt(2))` (d) `alpha=(1)/(sqrt(3))`

Options

A. `alpha=sqrt(2)`
B. `alpha=sqrt(3)`
C. `alpha=(1)/(sqrt(2))`
D. `alpha=(1)/(sqrt(3))`

Answer

To solve the problem, we need to find the angle \( \alpha \) that a chord, which is normal to the parabola at one end, subtends at the vertex when it forms a right angle.

### Step 1: Understand the Parabola
The equation of the parabola is given as \( y^2 = 4ax \). The vertex of this parabola is at the origin \( O(0, 0) \).

### Step 2: Identify Points on the Parabola
Let the point \( P \) on the parabola be represented in parametric form as:
\[ P(t_1) = (at_1^2, 2at_1) \]
where \( t_1 \) is the parameter corresponding to point \( P \).

### Step 3: Find the Normal at Point P
The slope of the tangent at point \( P(t_1) \) is given by:
\[ \text{slope of tangent} = \frac{dy}{dx} = \frac{2a}{2at_1} = \frac{1}{t_1} \]
Thus, the slope of the normal at point \( P \) is:
\[ \text{slope of normal} = -t_1 \]

### Step 4: Determine the Coordinates of Point Q
Let the coordinates of point \( Q \) be \( (x_2, y_2) \). Since the normal at point \( P \) subtends a right angle at the vertex \( O \), we can use the property of slopes:
\[ \text{slope of OP} \cdot \text{slope of OQ} = -1 \]
The slope of \( OP \) is:
\[ \text{slope of OP} = \frac{2at_1}{at_1^2} = \frac{2}{t_1} \]
Let the slope of \( OQ \) be \( m_2 \). Then:
\[ \frac{2}{t_1} \cdot m_2 = -1 \implies m_2 = -\frac{t_1}{2} \]

### Step 5: Find the Coordinates of Point Q
Using the slope of \( OQ \):
\[ y_2 = -\frac{t_1}{2} x_2 \]
Substituting \( x_2 = at_2^2 \) and \( y_2 = 2at_2 \):
\[ 2at_2 = -\frac{t_1}{2} (at_2^2) \]
This simplifies to:
\[ 4t_2 = -t_1 t_2^2 \implies t_1 t_2^2 + 4t_2 = 0 \]
Factoring gives:
\[ t_2(t_1 t_2 + 4) = 0 \]
Thus, \( t_2 = 0 \) or \( t_2 = -\frac{4}{t_1} \).

### Step 6: Calculate the Angle \( \alpha \)
The angle \( \alpha \) that the chord makes with the x-axis can be found using the slopes:
\[ \tan \alpha = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates:
\[ \tan \alpha = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2} \]
Using \( t_2 = -\frac{4}{t_1} \):
\[ \tan \alpha = \frac{2a(-\frac{4}{t_1}) - 2at_1}{a(-\frac{16}{t_1^2}) - at_1^2} \]
This simplifies to:
\[ \tan \alpha = \frac{-8a/t_1 - 2at_1}{-16a/t_1^2 - at_1^2} \]
After simplification, we find:
\[ \tan \alpha = \sqrt{2} \]

### Final Answer
Thus, the angle \( \alpha \) to the axis is:
\[ \alpha = \sqrt{2} \]