Find the angle at which normal at point `P(a t^2,2a t)` to the parabola meets the parabola again at point `Qdot`

Explanation

To solve the problem of finding the angle at which the normal at point \( P(at^2, 2at) \) to the parabola \( y^2 = 4ax \) meets the parabola again at point \( Q \), we can follow these steps:

### Step 1: Identify the point on the parabola
The point \( P \) on the parabola is given as \( P(at^2, 2at) \).

### Step 2: Find the slope of the normal at point \( P \)
For the parabola \( y^2 = 4ax \), the slope of the tangent at point \( P \) can be derived from the parametric equations. The slope of the tangent at \( P \) is given by:
\[ m_t = \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t} \]
Thus, the slope of the normal at point \( P \) is:
\[ m_n = -\frac{1}{m_t} = -t \]

### Step 3: Find the equation of the normal at point \( P \)
The equation of the normal at point \( P \) can be written using the point-slope form:
\[ y - 2at = -t(x - at^2) \]
Rearranging gives:
\[ y = -tx + at^2 + 2at \]
This is the equation of the normal line.

### Step 4: Find the intersection of the normal with the parabola
To find the point \( Q \) where the normal intersects the parabola again, we substitute the equation of the normal into the parabola's equation \( y^2 = 4ax \):
\[ (-tx + at^2 + 2at)^2 = 4ax \]
Expanding this and rearranging will lead to a quadratic equation in \( x \).

### Step 5: Solve for \( x \)
Let’s denote \( y = -tx + at^2 + 2at \) and substitute it into \( y^2 = 4ax \):
\[ (-tx + at^2 + 2at)^2 = 4ax \]
This will yield a quadratic equation in \( x \). The solutions will give the \( x \)-coordinates of the intersection points.

### Step 6: Find the slope of the tangent at point \( Q \)
Let’s denote the parameter corresponding to point \( Q \) as \( t_1 \). The slope of the tangent at point \( Q \) is:
\[ m_t' = \frac{1}{t_1} \]

### Step 7: Calculate the angle between the normal and the tangent
The angle \( \theta \) between the normal at point \( P \) and the tangent at point \( Q \) can be calculated using the formula:
\[ \tan \theta = \left| \frac{m_t' - m_n}{1 + m_t' m_n} \right| \]
Substituting \( m_n = -t \) and \( m_t' = \frac{1}{t_1} \):
\[ \tan \theta = \left| \frac{\frac{1}{t_1} + t}{1 - t \cdot \frac{1}{t_1}} \right| \]

### Step 8: Simplify and find \( \theta \)
After simplification, we can express \( \tan \theta \) in terms of \( t \) and \( t_1 \). Finally, we can find \( \theta \) by taking the arctangent:
\[ \theta = \tan^{-1} \left( \frac{|t|}{2} \right) \]

### Final Answer
The angle at which the normal at point \( P \) meets the parabola again at point \( Q \) is:
\[ \theta = \tan^{-1} \left( \frac{|t|}{2} \right) \]