Find the locus of point P, if two tangents are drawn from it to the parabola `y ^(2) = 4x` such that the slope of one tangent is three times the slope of other.

Answer

To find the locus of point P from which two tangents are drawn to the parabola \( y^2 = 4x \) such that the slope of one tangent is three times the slope of the other, we can follow these steps:

### Step 1: Understand the Parabola
The equation of the parabola is given by:
\[ y^2 = 4x \]
This is a standard parabola that opens to the right.

### Step 2: Define the Slopes of the Tangents
Let the slopes of the two tangents be \( m_1 \) and \( m_2 \). According to the problem, we have:
\[ m_1 = 3m_2 \]

### Step 3: Use the Tangent Equation
The equation of the tangent to the parabola \( y^2 = 4x \) at a point where the slope is \( m \) can be expressed as:
\[ y = mx - 2m^2 \]

### Step 4: Set Up the Slopes
From the tangent equation, we can express the slopes:
- For the first tangent with slope \( m_1 \):
\[ y = m_1 x - 2m_1^2 \]
- For the second tangent with slope \( m_2 \):
\[ y = m_2 x - 2m_2^2 \]

### Step 5: Relate the Slopes
Substituting \( m_1 = 3m_2 \) into the first tangent equation gives:
\[ y = 3m_2 x - 2(3m_2)^2 = 3m_2 x - 18m_2^2 \]
Now we have two equations:
1. \( y = 3m_2 x - 18m_2^2 \)
2. \( y = m_2 x - 2m_2^2 \)

### Step 6: Find the Intersection Point
To find the locus of point P, we need to find the intersection of these two tangents. Setting the two equations equal to each other:
\[ 3m_2 x - 18m_2^2 = m_2 x - 2m_2^2 \]
Rearranging gives:
\[ (3m_2 - m_2)x = 18m_2^2 - 2m_2^2 \]
\[ 2m_2 x = 16m_2^2 \]
Assuming \( m_2 \neq 0 \) (since if \( m_2 = 0 \), the tangents would be horizontal), we can divide both sides by \( m_2 \):
\[ 2x = 16m_2 \implies x = 8m_2 \]

### Step 7: Substitute Back to Find y
Substituting \( x = 8m_2 \) back into one of the tangent equations to find y:
Using \( y = m_2 x - 2m_2^2 \):
\[ y = m_2(8m_2) - 2m_2^2 = 8m_2^2 - 2m_2^2 = 6m_2^2 \]

### Step 8: Express in Terms of x and y
Now we have \( x = 8m_2 \) and \( y = 6m_2^2 \). We can express \( m_2 \) in terms of \( x \):
\[ m_2 = \frac{x}{8} \]
Substituting this into the equation for y:
\[ y = 6\left(\frac{x}{8}\right)^2 = 6 \cdot \frac{x^2}{64} = \frac{3x^2}{32} \]

### Step 9: Rearranging to Find the Locus
Rearranging gives:
\[ 3x^2 - 32y = 0 \]
or
\[ 3x^2 = 32y \]

### Final Result
Thus, the locus of point P is given by the equation:
\[ \boxed{3x^2 = 32y} \]