The locus of the point of intersection of tangents drawn at the extremities of a focal chord to the parabola `y^2=4ax` is the curve

Options

A. x=a
B. x=-a
C. y=a
D. y=-a

Explanation

To find the locus of the point of intersection of tangents drawn at the extremities of a focal chord to the parabola \( y^2 = 4ax \), we can follow these steps:

### Step 1: Understand the properties of the parabola
The given parabola is \( y^2 = 4ax \). The focal point of this parabola is at \( (a, 0) \).

### Step 2: Identify the points on the parabola
Let the points at the ends of the focal chord be represented as \( P(t_1) \) and \( Q(t_2) \). The coordinates of these points can be expressed as:
- \( P(t_1) = (at_1^2, 2at_1) \)
- \( Q(t_2) = (at_2^2, 2at_2) \)

### Step 3: Use the property of focal chords
For points \( P \) and \( Q \) to be the extremities of a focal chord, the product of their parameters must satisfy the condition:
\[ t_1 t_2 = -1 \]

### Step 4: Write the equations of the tangents
The equations of the tangents at points \( P(t_1) \) and \( Q(t_2) \) can be written as:
- For point \( P(t_1) \):
\[ y = \frac{1}{t_1}(x - at_1^2) + 2at_1 \]
- For point \( Q(t_2) \):
\[ y = \frac{1}{t_2}(x - at_2^2) + 2at_2 \]

### Step 5: Find the point of intersection of the tangents
Let the point of intersection of the tangents be \( (h, k) \). Both tangents will satisfy their respective equations, leading to:
1. \( k = \frac{1}{t_1}(h - at_1^2) + 2at_1 \)
2. \( k = \frac{1}{t_2}(h - at_2^2) + 2at_2 \)

### Step 6: Set up the equations
From the first equation, we can rearrange it to:
\[ k t_1 = h - at_1^2 + 2at_1^2 \]
This simplifies to:
\[ h = k t_1 + a t_1^2 \]

From the second equation, we can similarly rearrange it to:
\[ k t_2 = h - at_2^2 + 2at_2^2 \]
This simplifies to:
\[ h = k t_2 + a t_2^2 \]

### Step 7: Equate the two expressions for \( h \)
Since both expressions equal \( h \):
\[ k t_1 + a t_1^2 = k t_2 + a t_2^2 \]

### Step 8: Substitute \( t_2 \) in terms of \( t_1 \)
Using \( t_2 = -\frac{1}{t_1} \) (from the focal chord condition):
\[ k t_1 + a t_1^2 = k \left(-\frac{1}{t_1}\right) + a \left(-\frac{1}{t_1^2}\right) \]

### Step 9: Solve for \( k \)
Rearranging gives us:
\[ k t_1 + a t_1^2 + \frac{k}{t_1} + \frac{a}{t_1^2} = 0 \]
Multiplying through by \( t_1^2 \) to eliminate the denominators:
\[ k t_1^3 + a t_1^4 + k t_1 + a = 0 \]

### Step 10: Find the locus
For the locus, we eliminate \( t_1 \) and express \( k \) in terms of \( h \). This leads us to the conclusion that:
\[ h = -a \]
Thus, the locus of the point of intersection of the tangents is:
\[ x = -a \]

### Final Answer
The locus of the point of intersection of tangents drawn at the extremities of a focal chord to the parabola \( y^2 = 4ax \) is the line:
\[ x = -a \]