From an external point P, tangents are drawn to the parabola. Find the equation of the locus of P when these tangents make angles `theta_(1)andtheta_(2)` with the axis of the parabola such that `costheta_(1)costheta_(2)=mu`, where `mu` is a constant.

Answer

To find the equation of the locus of point P from which tangents are drawn to the parabola \( y^2 = 4Ax \), and where the tangents make angles \( \theta_1 \) and \( \theta_2 \) with the axis of the parabola such that \( \cos \theta_1 \cos \theta_2 = \mu \) (where \( \mu \) is a constant), we can follow these steps:

### Step 1: Set up the problem
Let the coordinates of point P be \( P(h, k) \). The parabola given is \( y^2 = 4Ax \).

### Step 2: Use the properties of tangents
For the parabola \( y^2 = 4Ax \), the equation of the tangent at a point \( (At^2, 2At) \) is given by:
\[ yy_1 = 2A(x + x_1) \]
where \( (x_1, y_1) \) is the point on the parabola.

### Step 3: Find the slopes of the tangents
The slopes of the tangents at points \( Q \) and \( R \) on the parabola can be expressed in terms of the parameters \( t_1 \) and \( t_2 \):
\[ m_1 = \frac{2A}{y_1} = \frac{2A}{2At_1} = \frac{1}{t_1}, \quad m_2 = \frac{2A}{y_2} = \frac{1}{t_2} \]

### Step 4: Relate angles to slopes
The angles \( \theta_1 \) and \( \theta_2 \) that the tangents make with the x-axis can be expressed as:
\[ \tan \theta_1 = \frac{1}{t_1}, \quad \tan \theta_2 = \frac{1}{t_2} \]
Thus, we have:
\[ \cos \theta_1 = \frac{1}{\sqrt{1 + \tan^2 \theta_1}} = \frac{t_1}{\sqrt{1 + t_1^2}}, \quad \cos \theta_2 = \frac{t_2}{\sqrt{1 + t_2^2}} \]

### Step 5: Substitute into the given condition
From the problem, we know:
\[ \cos \theta_1 \cos \theta_2 = \mu \]
Substituting the expressions for \( \cos \theta_1 \) and \( \cos \theta_2 \):
\[ \frac{t_1 t_2}{\sqrt{(1 + t_1^2)(1 + t_2^2)}} = \mu \]

### Step 6: Square both sides
Squaring both sides gives:
\[ \frac{t_1^2 t_2^2}{(1 + t_1^2)(1 + t_2^2)} = \mu^2 \]
Cross-multiplying yields:
\[ t_1^2 t_2^2 = \mu^2 (1 + t_1^2)(1 + t_2^2) \]

### Step 7: Expand and rearrange
Expanding the right-hand side:
\[ t_1^2 t_2^2 = \mu^2 + \mu^2 t_1^2 + \mu^2 t_2^2 + \mu^2 t_1^2 t_2^2 \]
Rearranging gives:
\[ (1 - \mu^2) t_1^2 t_2^2 - \mu^2 t_1^2 - \mu^2 t_2^2 - \mu^2 = 0 \]

### Step 8: Substitute \( h \) and \( k \)
Recall that:
\[ h = At_1 t_2, \quad k = A(t_1 + t_2) \]
Substituting these into the equation gives a relation between \( h \) and \( k \).

### Step 9: Final equation of the locus
After simplification, we arrive at the locus of point P:
\[ (1 - \mu^2) \frac{h^2}{A^2} - \mu^2 \frac{k^2}{A^2} + 2\mu^2 \frac{h}{A} - \mu^2 = 0 \]

This is the equation of the locus of point P.