The locus of point of intersection of tangents to `y^(2) = 4ax` which includes an angle `alpha` is

Options

A. `y^(2)-4ax=Tan^(2) alpha`
B. `y^(2)-4ax=(x+a)^(2)Cot^(2)alpha`
C. `y^(2)-4ax=(x+a)^(2) Tan^(2)alpha`
D. `y^(2)-4ax=(x-a)^(2)Cot^(2) alpha`

Answer

To find the locus of the point of intersection of tangents to the parabola \( y^2 = 4ax \) that include an angle \( \alpha \), we can follow these steps:

### Step 1: Define the Tangents
Let the points of tangency on the parabola be \( T_1(t_1) \) and \( T_2(t_2) \). The equations of the tangents at these points can be written as:
\[ T_1: t_1y = x + at_1^2 \]
\[ T_2: t_2y = x + at_2^2 \]

### Step 2: Find the Point of Intersection
Let the point of intersection of these tangents be \( (h, k) \). From the equations of the tangents, we can express \( h \) and \( k \) in terms of \( t_1 \) and \( t_2 \):
\[ h = at_1t_2 \]
\[ k = a(t_1 + t_2) \]

### Step 3: Relate the Slopes to the Angle
The slopes of the tangents are given by:
\[ \text{slope of } T_1 = \frac{1}{t_1}, \quad \text{slope of } T_2 = \frac{1}{t_2} \]
The angle \( \alpha \) between the two tangents can be expressed using the tangent of the angle:
\[ \tan \alpha = \frac{\frac{1}{t_2} - \frac{1}{t_1}}{1 + \frac{1}{t_1} \cdot \frac{1}{t_2}} = \frac{t_1 - t_2}{t_1t_2 + 1} \]

### Step 4: Square Both Sides
Squaring both sides gives:
\[ \tan^2 \alpha = \frac{(t_1 - t_2)^2}{(1 + t_1t_2)^2} \]

### Step 5: Cross Multiply
Cross-multiplying leads to:
\[ (1 + t_1t_2)^2 \tan^2 \alpha = (t_1 - t_2)^2 \]

### Step 6: Expand the Right Side
Expanding the right side:
\[ (t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1t_2 \]

### Step 7: Substitute for \( t_1 + t_2 \) and \( t_1t_2 \)
Using the expressions for \( t_1 + t_2 \) and \( t_1t_2 \):
\[ t_1 + t_2 = \frac{k}{a}, \quad t_1t_2 = \frac{h}{a} \]
Substituting these into the equation gives:
\[ \left(1 + \frac{h}{a}\right)^2 \tan^2 \alpha = \left(\frac{k}{a}\right)^2 - 4\frac{h}{a} \]

### Step 8: Rearranging the Equation
After simplifying and rearranging, we find:
\[ a + h \tan^2 \alpha = k^2 - 4ah \]

### Step 9: Final Locus Equation
Substituting \( h \) and \( k \) back into the equation gives us the locus:
\[ x + a \tan^2 \alpha = y^2 - 4ax \]

### Conclusion
Thus, the locus of the point of intersection of the tangents is given by:
\[ x + a = \frac{y^2 - 4ax}{\tan^2 \alpha} \]