Let the tangents to the parabola `y^(2)=4ax` drawn from point P have slope `m_(1) and m_(2)`. If `m_(1)m_(2)=2`, then the locus of point P is

Options

A. `x=a`
B. `x=(a)/(2)`
C. `x+a=0`
D. `x=2a`

Answer

To solve the problem, we need to find the locus of the point P from which tangents to the parabola \( y^2 = 4ax \) can be drawn, given that the product of the slopes of these tangents is equal to 2.

### Step-by-Step Solution:

1. **Identify the Point P**:
Let the coordinates of point P be \( (x_1, y_1) \).

2. **Equation of Tangent**:
The equation of the tangent to the parabola \( y^2 = 4ax \) at a point with slope \( m \) is given by:
\[ y = mx + \frac{a}{m} \]

3. **Substituting Point P into the Tangent Equation**:
Since point P lies on the tangent, we substitute \( (x_1, y_1) \) into the tangent equation:
\[ y_1 = mx_1 + \frac{a}{m} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ mx_1 - y_1 + \frac{a}{m} = 0 \]
Multiplying through by \( m \) (to eliminate the fraction) gives:
\[ mx_1 - y_1 m + a = 0 \]

5. **Forming a Quadratic in m**:
Rearranging the above equation leads to:
\[ m^2 x_1 - y_1 m + a = 0 \]
This is a quadratic equation in \( m \).

6. **Using the Product of Roots**:
The product of the roots \( m_1 \) and \( m_2 \) of the quadratic equation \( am^2 + bm + c = 0 \) is given by:
\[ m_1 m_2 = \frac{c}{a} \]
In our case:
\[ m_1 m_2 = \frac{a}{x_1} \]
We are given that \( m_1 m_2 = 2 \). Thus, we have:
\[ \frac{a}{x_1} = 2 \]

7. **Solving for x_1**:
From the equation above, we can solve for \( x_1 \):
\[ x_1 = \frac{a}{2} \]

8. **Finding the Locus**:
The locus of point P, as \( a \) varies, can be expressed by replacing \( x_1 \) with \( x \):
\[ x = \frac{a}{2} \]
Rearranging gives:
\[ a = 2x \]

### Final Answer:
The locus of point P is given by the equation:
\[ x = \frac{a}{2} \]