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 solve the problem step by step, we will find the locus of point P from which two tangents are drawn to the parabola \( y^2 = 4ax \), given that the slope of one tangent is twice the slope of the other.

### 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:
\[ y = mx + \frac{a}{m} \]

### Step 3: Set up the slopes of the tangents
Let the slope of the first tangent be \( m_1 \). Then, the slope of the second tangent will be \( m_2 = 2m_1 \).

### Step 4: Substitute the coordinates of point P into the tangent equation
Since point P lies on both tangents, we can substitute \( (h, k) \) into the tangent equations:
1. For the first tangent:
\[ k = m_1 h + \frac{a}{m_1} \]
2. For the second tangent:
\[ k = 2m_1 h + \frac{a}{2m_1} \]

### Step 5: Rearranging the equations
Rearranging both equations gives us:
1. \( k m_1 = h m_1^2 + a \) (Equation 1)
2. \( k = 2m_1 h + \frac{a}{2m_1} \) (Equation 2)

### Step 6: Form a quadratic equation
From Equation 1, we can express it as:
\[ h m_1^2 - k m_1 + a = 0 \]
This is a quadratic equation in \( m_1 \).

### Step 7: Roots of the quadratic equation
The roots of this quadratic equation are \( m_1 \) and \( 2m_1 \). For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots \( (m_1 + 2m_1) \) is given by:
\[ m_1 + 2m_1 = -\frac{-k}{h} = \frac{k}{h} \]
Thus,
\[ 3m_1 = \frac{k}{h} \quad \Rightarrow \quad m_1 = \frac{k}{3h} \]

### Step 8: Product of the roots
The product of the roots \( (m_1 \cdot 2m_1) \) is given by:
\[ m_1 \cdot 2m_1 = \frac{a}{h} \quad \Rightarrow \quad 2m_1^2 = \frac{a}{h} \]
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: Cross-multiplying
Cross-multiplying gives:
\[ 2k^2 = 9ah \]

### Step 10: Finding the locus
Rearranging gives:
\[ k^2 = \frac{9}{2}ah \]
Replacing \( h \) with \( x \) and \( k \) with \( y \) (since \( h \) and \( k \) represent the coordinates of point P):
\[ y^2 = \frac{9}{2}ax \]

### Conclusion
The locus of point P is given by the equation:
\[ y^2 = \frac{9}{2}ax \]
This represents a parabola.