From a point P, two tangents are drawn to the parabola `y^(2) = 4ax`. If the slope of one tagents is twice the slope of other, the locus of P is

Options

A. Circle
B. Straight line
C. Parabola
D. Ellipse

Answer

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

### Step 1: Define the point P
Let the coordinates of point P be \( (h, k) \).

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

### Step 3: Set up the relationship between slopes
Let the slope of the first tangent be \( m_1 \) and the slope of the second tangent be \( m_2 = 2m_1 \).

### Step 4: Substitute point P into the tangent equation
Substituting the coordinates of point P into the tangent equation gives:
\[ k = m_1 h + \frac{a}{m_1} \quad \text{(for the first tangent)} \]
\[ k = 2m_1 h + \frac{a}{2m_1} \quad \text{(for the second tangent)} \]

### Step 5: Rearrange the equations
Rearranging these equations, we get:
1. \( k m_1 = h m_1^2 + a \) (from the first tangent)
2. \( k m_2 = h m_2^2 + a \) (from the second tangent)

### Step 6: Form a quadratic equation
From the first equation, we can express it as:
\[ h m_1^2 - k m_1 + a = 0 \]
This is a quadratic equation in \( m_1 \).

### Step 7: Use the properties of roots
Let the roots of this quadratic be \( m_1 \) and \( m_2 = 2m_1 \). By Vieta's formulas:
- The sum of the roots \( m_1 + m_2 = \frac{k}{h} \)
- The product of the roots \( m_1 \cdot m_2 = \frac{a}{h} \)

Substituting \( m_2 = 2m_1 \) into the sum gives:
\[ m_1 + 2m_1 = 3m_1 = \frac{k}{h} \]
Thus, we have:
\[ m_1 = \frac{k}{3h} \]

### Step 8: Substitute back to find \( m_1^2 \)
Now substituting \( m_1 \) back into the product of the roots:
\[ m_1 \cdot 2m_1 = 2m_1^2 = \frac{a}{h} \]
Substituting \( m_1 = \frac{k}{3h} \):
\[ 2\left(\frac{k}{3h}\right)^2 = \frac{a}{h} \]
This simplifies to:
\[ \frac{2k^2}{9h^2} = \frac{a}{h} \]

### Step 9: Rearranging the equation
Cross-multiplying gives:
\[ 2k^2 = 9ah \]

### Step 10: Express in terms of locus
Rearranging this gives:
\[ k^2 = \frac{9ah}{2} \]
Substituting back \( k = y \) and \( h = x \):
\[ y^2 = \frac{9}{2} ax \]

### Final Result
The locus of point P is:
\[ y^2 = \frac{9}{2} ax \]