prove that the locus of the point of intersection of the tangents at the extremities of any chord of the parabola `y^2 = 4ax` which subtends a right angle at the vertes is `x+4a=0`.

Answer

To prove that the locus of the point of intersection of the tangents at the extremities of any chord of the parabola \( y^2 = 4ax \) which subtends a right angle at the vertex is given by \( x + 4a = 0 \), we can follow these steps:

### Step 1: Understand the Parabola
The equation of the parabola is given by:
\[ y^2 = 4ax \]
This is a standard form of a parabola that opens to the right.

### Step 2: Define the Chord
Let \( A(t_1) \) and \( B(t_2) \) be the points on the parabola corresponding to parameters \( t_1 \) and \( t_2 \). The coordinates of these points can be expressed as:
\[ A(t_1) = (at_1^2, 2at_1) \quad \text{and} \quad B(t_2) = (at_2^2, 2at_2) \]

### Step 3: Condition for Right Angle
Since the chord \( AB \) subtends a right angle at the vertex, we have the condition:
\[ \tan(\theta_1) \cdot \tan(\theta_2) = -1 \]
where \( \theta_1 \) and \( \theta_2 \) are the angles of the tangents at points \( A \) and \( B \) with respect to the x-axis.

### Step 4: Find the Tangents
The equations of the tangents at points \( A \) and \( B \) can be derived using the formula for the tangent to the parabola:
\[ yy_1 = 2a(x + x_1) \]
Thus, the tangents at \( A(t_1) \) and \( B(t_2) \) are:
\[ yt_1 = 2a(x + at_1^2) \quad \text{(1)} \]
\[ yt_2 = 2a(x + at_2^2) \quad \text{(2)} \]

### Step 5: Find the Point of Intersection
To find the point of intersection \( P(h, k) \) of the two tangents, we can solve equations (1) and (2) simultaneously.

From (1):
\[ k = \frac{2a}{t_1}(h + at_1^2) \]
From (2):
\[ k = \frac{2a}{t_2}(h + at_2^2) \]

Equating the two expressions for \( k \):
\[ \frac{2a}{t_1}(h + at_1^2) = \frac{2a}{t_2}(h + at_2^2) \]

### Step 6: Simplify the Equation
Cancelling \( 2a \) from both sides (assuming \( a \neq 0 \)):
\[ \frac{1}{t_1}(h + at_1^2) = \frac{1}{t_2}(h + at_2^2) \]

Cross-multiplying gives:
\[ t_2(h + at_1^2) = t_1(h + at_2^2) \]

Expanding and rearranging:
\[ t_2h + at_2t_1^2 = t_1h + at_1t_2^2 \]
\[ (t_2 - t_1)h = a(t_1t_2^2 - t_2t_1^2) \]

### Step 7: Solve for \( h \)
Rearranging gives:
\[ h = \frac{a(t_1t_2^2 - t_2t_1^2)}{t_2 - t_1} \]
Factoring out \( t_1t_2 \):
\[ h = a(t_1 + t_2) \]

### Step 8: Find the Locus
Using the property of the tangents, we know that:
\[ h = -4a \]
Thus, the locus of the point of intersection of the tangents is:
\[ x + 4a = 0 \]

### Conclusion
Therefore, we have proved that the locus of the point of intersection of the tangents at the extremities of any chord of the parabola that subtends a right angle at the vertex is given by:
\[ \boxed{x + 4a = 0} \]