Let `L_(1),L_(2)andL_(3)` be the three normals to the parabola `y^(2)=4ax` from point P inclined at the angle `theta_(1),theta_(2)andtheta_(3)` with x-axis, respectively. Then find the locus of point P given that `theta_(1)+theta_(2)+theta_(3)=alpha` (constant).

Explanation

To find the locus of the point P from which three normals to the parabola \( y^2 = 4ax \) are drawn, and given that the angles of inclination of these normals with the x-axis are \( \theta_1, \theta_2, \theta_3 \) such that \( \theta_1 + \theta_2 + \theta_3 = \alpha \) (a constant), we can follow these steps:

### Step 1: Understand the Equation of the Normal
The equation of the normal to the parabola \( y^2 = 4ax \) at a point with slope \( m \) is given by:
\[ y = mx - 2am - a m^3 \]

### Step 2: Substitute the Point P
Let the point P be \( (h, k) \). The normal passing through P can be expressed as:
\[ k = mh - 2am - am^3 \]

### Step 3: Rearranging the Equation
Rearranging the equation gives us:
\[ am^3 + 2am + (k - mh) = 0 \]
This is a cubic equation in terms of \( m \).

### Step 4: Roots of the Cubic Equation
Let the roots of this cubic equation be \( m_1, m_2, m_3 \) corresponding to slopes \( \tan \theta_1, \tan \theta_2, \tan \theta_3 \). By Vieta's formulas, we know:
1. \( m_1 + m_2 + m_3 = 0 \)
2. \( m_1 m_2 + m_2 m_3 + m_3 m_1 = \frac{2a - h}{a} \)
3. \( m_1 m_2 m_3 = -\frac{k}{a} \)

### Step 5: Use the Given Condition
Since \( \theta_1 + \theta_2 + \theta_3 = \alpha \), we can express this in terms of tangent:
\[ \tan(\theta_1 + \theta_2 + \theta_3) = \tan(\alpha) \]
Using the tangent addition formula:
\[ \tan(\theta_1 + \theta_2 + \theta_3) = \frac{\tan \theta_1 + \tan \theta_2 + \tan \theta_3 - \tan \theta_1 \tan \theta_2 \tan \theta_3}{1 - (\tan \theta_1 \tan \theta_2 + \tan \theta_2 \tan \theta_3 + \tan \theta_3 \tan \theta_1)} \]

### Step 6: Substitute the Values
Substituting \( m_1, m_2, m_3 \) into the equation gives:
\[ \tan \alpha = \frac{0 - (-\frac{k}{a})}{1 - \frac{2a - h}{a}} \]
This simplifies to:
\[ \tan \alpha = \frac{\frac{k}{a}}{1 - \frac{2a - h}{a}} = \frac{\frac{k}{a}}{\frac{h}{a}} = \frac{k}{h} \]

### Step 7: Final Locus Equation
Thus, we have:
\[ k = h \tan \alpha \]
Replacing \( k \) with \( y \) and \( h \) with \( x \), we get the locus of point P:
\[ y = x \tan \alpha \]

### Conclusion
The locus of the point P is a straight line given by:
\[ y = x \tan \alpha \]