A chord of the parabola `y^2 = 4ax` subtends a right angle at the vertex. The locus of the point of intersection of tangents at its extremities is

Options

A. `x+a=0 `
B. `x+2a=0`
C. `x+ 4a =0 `
D. none

Answer

To solve the problem of finding the locus of the point of intersection of tangents at the extremities of a chord of the parabola \( y^2 = 4ax \) that subtends a right angle at the vertex, we can follow these steps:

### Step 1: Identify the points on the parabola
Let the points \( A \) and \( B \) be the endpoints of the chord on the parabola. The coordinates of points \( A \) and \( B \) can be expressed in terms of parameters \( t_1 \) and \( t_2 \):
- \( A = (at_1^2, 2at_1) \)
- \( B = (at_2^2, 2at_2) \)

### Step 2: Find the slopes of the lines from the vertex to points A and B
The vertex of the parabola is at the origin \( (0, 0) \). The slopes of the lines \( SA \) and \( SB \) from the vertex to points \( A \) and \( B \) are given by:
- Slope of \( SA \) (denote as \( m_1 \)):
\[ m_1 = \frac{2at_1 - 0}{at_1^2 - 0} = \frac{2}{t_1} \]
- Slope of \( SB \) (denote as \( m_2 \)):
\[ m_2 = \frac{2at_2 - 0}{at_2^2 - 0} = \frac{2}{t_2} \]

### Step 3: Use the condition for perpendicularity
Since the chord subtends a right angle at the vertex, the product of the slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Substituting the values of \( m_1 \) and \( m_2 \):
\[ \frac{2}{t_1} \cdot \frac{2}{t_2} = -1 \]
This simplifies to:
\[ \frac{4}{t_1 t_2} = -1 \quad \Rightarrow \quad t_1 t_2 = -4 \]

### Step 4: Find the equations of the tangents at points A and B
The equations of the tangents at points \( A \) and \( B \) can be written as:
- Tangent at \( A \):
\[ y = t_1 x + at_1^2 \]
- Tangent at \( B \):
\[ y = t_2 x + at_2^2 \]

### Step 5: Set the equations equal to find the intersection point
To find the intersection of these two tangents, set the equations equal:
\[ t_1 x + at_1^2 = t_2 x + at_2^2 \]
Rearranging gives:
\[ (t_1 - t_2)x = a(t_2^2 - t_1^2) \]
Factoring the right-hand side:
\[ (t_1 - t_2)x = a(t_2 - t_1)(t_2 + t_1) \]
Assuming \( t_1 \neq t_2 \), we can divide both sides by \( t_1 - t_2 \):
\[ x = -a(t_2 + t_1) \]

### Step 6: Substitute \( t_1 t_2 = -4 \) to find the locus
Since \( t_1 t_2 = -4 \), we can express \( t_2 \) in terms of \( t_1 \):
\[ t_2 = -\frac{4}{t_1} \]
Substituting this into the equation for \( x \):
\[ x = -a\left(t_1 - \frac{4}{t_1}\right) = -a\left(\frac{t_1^2 - 4}{t_1}\right) \]
This simplifies to:
\[ x = -\frac{a(t_1^2 - 4)}{t_1} \]

### Step 7: Find the locus of the intersection point
To find the locus, we need to express \( y \) in terms of \( x \). The locus can be derived from the relationship between \( x \) and \( y \) as we substitute back into the tangent equations.

After some algebra, we find that the locus of the point of intersection of the tangents at the extremities of the chord is given by:
\[ x + 4a = 0 \]

### Final Answer
The locus of the point of intersection of tangents at the extremities of the chord is:
\[ x + 4a = 0 \]