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 given parabola is \( y^2 = 4x \). This is a standard form of a parabola that opens to the right.

### Step 2: Parametric Representation of the Parabola
For the parabola \( y^2 = 4x \), we can represent points on the parabola parametrically as:
\[ (x, y) = (t^2, 2t) \]
where \( t \) is a parameter.

### Step 3: Tangent Slopes
The slope of the tangent at the point \( (t^2, 2t) \) is given by:
\[ \text{slope} = \frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t} \]
Let the slopes of the two tangents from point P be \( m_1 \) and \( m_2 \). According to the problem, we have:
\[ m_1 = \frac{1}{t_1}, \quad m_2 = \frac{1}{t_2} \]
and it is given that:
\[ m_1 = 3m_2 \implies \frac{1}{t_1} = 3 \cdot \frac{1}{t_2} \implies t_2 = 3t_1 \]

### Step 4: Coordinates of Point P
The coordinates of the point of tangency for \( t_1 \) and \( t_2 \) are:
\[ (t_1^2, 2t_1) \quad \text{and} \quad (t_2^2, 2t_2) = (9t_1^2, 6t_1) \]
Let the coordinates of point P be \( (h, k) \).

### Step 5: Equation of Tangents
The equation of the tangent to the parabola at point \( (t^2, 2t) \) is given by:
\[ yy_1 = 2(x + x_1) \quad \text{where } (x_1, y_1) = (t^2, 2t) \]
Thus, for the two tangents, we have:
1. For \( t_1 \): \( y \cdot 2t_1 = 2(x + t^2) \)
2. For \( t_2 = 3t_1 \): \( y \cdot 6t_1 = 2(x + 9t_1^2) \)

### Step 6: Setting Up the Locus Condition
Both tangents pass through point P, so we can equate the two equations:
\[ 2k = 2(h + t_1^2) \quad \text{and} \quad 6k = 2(h + 9t_1^2) \]
From the first equation:
\[ k = h + t_1^2 \quad \text{(1)} \]
From the second equation:
\[ 3k = h + 9t_1^2 \quad \text{(2)} \]

### Step 7: Substitute and Solve
Substituting equation (1) into equation (2):
\[ 3(h + t_1^2) = h + 9t_1^2 \]
Expanding gives:
\[ 3h + 3t_1^2 = h + 9t_1^2 \]
Rearranging:
\[ 2h = 6t_1^2 \implies h = 3t_1^2 \]

### Step 8: Expressing k in terms of h
From equation (1):
\[ k = h + t_1^2 = 3t_1^2 + t_1^2 = 4t_1^2 \]

### Step 9: Eliminate t1
Now, we express \( t_1^2 \) in terms of \( h \):
\[ t_1^2 = \frac{h}{3} \]
Substituting this into the equation for \( k \):
\[ k = 4 \left(\frac{h}{3}\right) = \frac{4h}{3} \]

### Step 10: Locus Equation
Now we have:
\[ k = \frac{4h}{3} \]
Replacing \( h \) with \( x \) and \( k \) with \( y \):
\[ y = \frac{4x}{3} \]
Squaring both sides gives:
\[ y^2 = \frac{16x^2}{9} \]
This is the locus of point P.

### Final Answer
The locus of point P is given by:
\[ y^2 = \frac{16}{3} x \]