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 find the locus of the point P from which tangents are drawn to the parabola \( y^2 = 4ax \) with the condition that the product of the slopes of the tangents \( m_1 \) and \( m_2 \) is equal to 2, we can follow these steps:

### Step 1: Equation of the Tangent
The equation of the tangent to the parabola \( y^2 = 4ax \) at a point with slope \( m \) is given by:
\[ y = mx + \frac{a}{m} \]

### Step 2: Point P Coordinates
Let the coordinates of point P be \( (h, k) \). The point P lies on the tangent line, so substituting \( y = k \) and \( x = h \) into the tangent equation gives:
\[ k = mh + \frac{a}{m} \]

### Step 3: Rearranging the Equation
Rearranging this equation leads to:
\[ k - mh - \frac{a}{m} = 0 \]
Multiplying through by \( m \) to eliminate the fraction results in:
\[ km - m^2h - a = 0 \]

### Step 4: Forming a Quadratic Equation
This is a quadratic equation in terms of \( m \):
\[ -m^2h + km - a = 0 \]
From the standard form \( Am^2 + Bm + C = 0 \), we identify:
- \( A = -h \)
- \( B = k \)
- \( C = -a \)

### Step 5: Condition for Slopes
The product of the slopes \( m_1 \) and \( m_2 \) from the quadratic equation is given by:
\[ m_1 m_2 = \frac{C}{A} = \frac{-a}{-h} = \frac{a}{h} \]
According to the problem, we have:
\[ m_1 m_2 = 2 \]
Thus:
\[ \frac{a}{h} = 2 \implies h = \frac{a}{2} \]

### Step 6: Locus of Point P
Since \( h = \frac{a}{2} \), we can express this in terms of the coordinates of point P:
\[ x = \frac{a}{2} \]
This indicates that the locus of point P is a vertical line given by:
\[ x = \frac{a}{2} \]

### Final Answer
The locus of point P is:
\[ \text{Locus: } x = \frac{a}{2} \]