The locus of the point through which pass three normals to the parabola `y^2=4ax`, such that two of them make angles `alpha&beta` respectively with the axis & `tanalpha *tanbeta = 2` is` (a > 0)`

Answer

To solve the problem, we need to find the locus of a point from which three normals to the parabola \( y^2 = 4ax \) pass, given that two of these normals make angles \( \alpha \) and \( \beta \) with the x-axis such that \( \tan \alpha \tan \beta = 2 \).

### Step-by-Step Solution:

1. **Understanding the Parabola**: The given parabola is \( y^2 = 4ax \). A point on this parabola can be represented as \( (at^2, 2at) \) where \( t \) is a parameter.

2. **Normal to the Parabola**: The equation of the normal to the parabola at the point \( (at^2, 2at) \) is given by:
\[ y - 2at = -\frac{1}{t}(x - at^2) \]
Rearranging this, we get:
\[ y = -\frac{1}{t}x + \left(2at + \frac{at^2}{t}\right) = -\frac{1}{t}x + 3a \]

3. **Slope of the Normal**: The slope \( m \) of the normal is \( -\frac{1}{t} \). We denote the slopes of the two normals making angles \( \alpha \) and \( \beta \) with the x-axis as \( m_1 = \tan \alpha \) and \( m_2 = \tan \beta \).

4. **Using the Condition**: From the problem, we know that:
\[ m_1 m_2 = 2 \]
This implies:
\[ \tan \alpha \tan \beta = 2 \]

5. **Finding the Third Normal**: Let the slope of the third normal be \( m_3 \). The product of the slopes of the three normals can be expressed as:
\[ m_1 m_2 m_3 = -1 \]
Substituting \( m_1 m_2 = 2 \):
\[ 2m_3 = -1 \implies m_3 = -\frac{1}{2} \]

6. **Equation of the Third Normal**: The equation of the normal with slope \( m_3 = -\frac{1}{2} \) passing through the point \( (h, k) \) is:
\[ k - 2at = -\frac{1}{2}(h - at^2) \]
Rearranging gives:
\[ 2k - 4at = -h + at^2 \implies at^2 + h - 2k + 4at = 0 \]

7. **Finding the Locus**: The above equation is a quadratic in \( t \). For the normals to be real, the discriminant must be non-negative:
\[ \Delta = (4a)^2 - 4a(h - 2k) \geq 0 \]
Simplifying gives:
\[ 16a^2 - 4a(h - 2k) \geq 0 \implies 4a(4a - h + 2k) \geq 0 \]
Since \( a > 0 \), we can divide by \( 4a \):
\[ 4a - h + 2k \geq 0 \implies h - 2k \leq 4a \]

8. **Substituting for Locus**: Replacing \( h \) with \( x \) and \( k \) with \( y \), we get:
\[ x - 2y = 4a \]
Rearranging gives:
\[ y^2 = 4px \quad \text{where } p = a \]

### Final Answer:
The locus of the point \( (h, k) \) is given by:
\[ y^2 = 4px \]