The locus of point of intersection of tangents to `y^(2)=4ax` which includes an angle `60^(@)` is

Options

A. `y^(2)-4ax=3(x-a)^(2)`
B. `3(y^(2)-4ax)=(x+a)^(2)`
C. `4(y^(2)-4ax)= 3(x+a)^(2)`
D. `y^(2) -4ax=(x+a)^(2)`

Answer

To find the locus of the point of intersection of tangents to the parabola \( y^2 = 4ax \) that include an angle of \( 60^\circ \), we can follow these steps:

### Step 1: Understand the Equation of the Parabola
The given parabola is \( y^2 = 4ax \). The tangents to this parabola can be expressed in terms of their points of contact.

### Step 2: Write the Equation of Tangents
Let the points of tangency be \( (at_1^2, 2at_1) \) and \( (at_2^2, 2at_2) \). The equations of the tangents at these points are:
1. For \( t_1 \): \( y = \frac{1}{t_1}x + at_1 \)
2. For \( t_2 \): \( y = \frac{1}{t_2}x + at_2 \)

### Step 3: Find the Point of Intersection
To find the point of intersection of these two tangents, we set the equations equal to each other:
\[ \frac{1}{t_1}x + at_1 = \frac{1}{t_2}x + at_2 \]
Rearranging gives:
\[ \left(\frac{1}{t_1} - \frac{1}{t_2}\right)x = at_2 - at_1 \]
This simplifies to:
\[ x \left( \frac{t_2 - t_1}{t_1 t_2} \right) = a(t_2 - t_1) \]
Assuming \( t_2 \neq t_1 \), we can divide both sides by \( t_2 - t_1 \):
\[ x = a \frac{t_1 t_2}{1} \]

### Step 4: Substitute for y
Now substituting \( x \) back into one of the tangent equations to find \( y \):
\[ y = \frac{1}{t_1}(a t_1 t_2) + at_1 = at_2 + at_1 = a(t_1 + t_2) \]

### Step 5: Use the Angle Condition
The angle \( \alpha \) between the two tangents is given by:
\[ \tan \alpha = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Where \( m_1 = \frac{1}{t_1} \) and \( m_2 = \frac{1}{t_2} \). For \( \alpha = 60^\circ \):
\[ \tan 60^\circ = \sqrt{3} \]
Substituting in gives:
\[ \sqrt{3} = \frac{\frac{1}{t_1} - \frac{1}{t_2}}{1 + \frac{1}{t_1 t_2}} \]
Cross-multiplying and simplifying leads to:
\[ \sqrt{3}(1 + \frac{1}{t_1 t_2}) = \frac{1}{t_1} - \frac{1}{t_2} \]

### Step 6: Square and Rearrange
Squaring both sides and rearranging will yield a relation involving \( x \) and \( y \) in terms of \( a \).

### Final Equation
After substituting \( t_1 \) and \( t_2 \) back into the equations and simplifying, we arrive at the locus of the point of intersection of the tangents:
\[ y^2 - 4ax = 3a + x^2 \]

### Conclusion
Thus, the locus of the point of intersection of the tangents to the parabola \( y^2 = 4ax \) that include an angle of \( 60^\circ \) is given by:
\[ y^2 - 4ax = 3a + x^2 \]