Find the locus of the point of intersection of two tangents to the parabola `y^(2)=4ax`, which with the tangent at the vertex forms a triangle of constant area 4sq. units.

Answer

To find the locus of the point of intersection of two tangents to the parabola \( y^2 = 4ax \) that, along with the tangent at the vertex, forms a triangle of constant area \( 4 \) square units, we can follow these steps:

### Step 1: Understand the Parabola and Points
The parabola is given by the equation \( y^2 = 4ax \). The vertex of this parabola is at the origin \( (0, 0) \). We will consider two points \( P \) and \( Q \) on the parabola, which can be represented in parametric form as:
- \( P(aT_1^2, 2aT_1) \)
- \( Q(aT_2^2, 2aT_2) \)

### Step 2: Find the Tangent Equations
The equation of the tangent to the parabola at point \( P \) is given by:
\[ y = T_1x - aT_1^2 \]
Similarly, the equation of the tangent at point \( Q \) is:
\[ y = T_2x - aT_2^2 \]

### Step 3: Find the Intersection of the Tangents
To find the point of intersection \( R \) of the tangents at \( P \) and \( Q \), we set the two tangent equations equal to each other:
\[ T_1x - aT_1^2 = T_2x - aT_2^2 \]
Rearranging gives:
\[ (T_1 - T_2)x = a(T_1^2 - T_2^2) \]
Thus, we can solve for \( x \):
\[ x = \frac{a(T_1 + T_2)}{T_1 - T_2} \]

### Step 4: Find the \( y \)-coordinate of Intersection
Substituting \( x \) back into either tangent equation gives us the \( y \)-coordinate:
\[ y = T_1\left(\frac{a(T_1 + T_2)}{T_1 - T_2}\right) - aT_1^2 \]
Simplifying this expression gives:
\[ y = \frac{aT_1(T_1 + T_2) - aT_1^2(T_1 - T_2)}{T_1 - T_2} \]

### Step 5: Area of Triangle Formed
The area \( A \) of triangle formed by the tangents at \( P \), \( Q \), and the vertex \( (0, 0) \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Here, the base is the distance between the points \( P \) and \( Q \) on the x-axis, and the height is the y-coordinate of the intersection point \( R \).

### Step 6: Set Area to Constant
Given that the area is constant at \( 4 \) square units, we set up the equation:
\[ \frac{1}{2} \times \text{Base} \times \text{Height} = 4 \]
From this, we can derive a relationship between \( T_1 \) and \( T_2 \).

### Step 7: Locus of Point \( R \)
Assuming \( R \) has coordinates \( (h, k) \), we can express \( h \) and \( k \) in terms of \( T_1 \) and \( T_2 \):
\[ h = aT_1T_2, \quad k = a(T_1 + T_2) \]
Using these relationships, we can eliminate \( T_1 \) and \( T_2 \) to find the locus equation.

### Step 8: Final Locus Equation
After manipulation and substitution, we arrive at the locus equation:
\[ k^2 - 4ah = 64 \]
Replacing \( h \) with \( x \) and \( k \) with \( y \), we get:
\[ y^2 - 4ax = 64 \]