Two tangents to the parabola `y^2=4ax` make supplementary angles with the x-axis. Then the locus of their point of intersection is

Answer

To find the locus of the point of intersection of two tangents to the parabola \( y^2 = 4ax \) that make supplementary angles with the x-axis, we can follow these steps:

### Step 1: Determine the equations of the tangents
The equation of the tangent to the parabola \( y^2 = 4ax \) at a point \( (at^2, 2at) \) is given by:
\[ yt = x + at^2 \]
For two points on the parabola, let’s consider the parameters \( t_1 \) and \( t_2 \). The equations of the tangents at these points are:
1. For \( t_1 \):
\[ y t_1 = x + a t_1^2 \]
2. For \( t_2 \):
\[ y t_2 = x + a t_2^2 \]

### Step 2: Find the point of intersection of the tangents
To find the point of intersection of these two tangents, we can solve the two equations simultaneously. Rearranging both equations gives:
1. \( y = \frac{x + a t_1^2}{t_1} \)
2. \( y = \frac{x + a t_2^2}{t_2} \)

Setting these equal to each other:
\[ \frac{x + a t_1^2}{t_1} = \frac{x + a t_2^2}{t_2} \]

Cross-multiplying gives:
\[ t_2(x + a t_1^2) = t_1(x + a t_2^2) \]

Expanding both sides:
\[ t_2 x + a t_2 t_1^2 = t_1 x + a t_1 t_2^2 \]

Rearranging terms:
\[ (t_2 - t_1)x = a(t_1 t_2^2 - t_2 t_1^2) \]

### Step 3: Solve for x
If \( t_2 \neq t_1 \), we can solve for \( x \):
\[ x = \frac{a(t_1 t_2^2 - t_2 t_1^2)}{t_2 - t_1} \]

### Step 4: Use the condition of supplementary angles
Since the tangents make supplementary angles, we know:
\[ \theta_1 + \theta_2 = \pi \]
This implies:
\[ \tan(\theta_1 + \theta_2) = 0 \]
Using the tangent addition formula:
\[ \tan(\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} = 0 \]
This means:
\[ \tan \theta_1 + \tan \theta_2 = 0 \]
Substituting \( \tan \theta_1 = \frac{1}{t_1} \) and \( \tan \theta_2 = \frac{1}{t_2} \):
\[ \frac{1}{t_1} + \frac{1}{t_2} = 0 \implies t_1 + t_2 = 0 \]
Thus, \( t_2 = -t_1 \).

### Step 5: Substitute back to find the locus
Substituting \( t_2 = -t_1 \) into the expression for \( x \):
\[ x = \frac{a(t_1(-t_1^2) - (-t_1)t_1^2)}{(-t_1) - t_1} = \frac{a(-t_1^3 + t_1^3)}{-2t_1} = 0 \]
Thus, \( x = 0 \).

### Step 6: Find y-coordinate
Substituting \( t_2 = -t_1 \) into either tangent equation gives:
\[ y = \frac{0 + a t_1^2}{t_1} = a t_1 \]
Since \( t_1 \) can take any value, \( y \) can take any value as well.

### Conclusion
The locus of the point of intersection of the tangents is:
\[ x = 0 \quad \text{(the y-axis)} \]