Any two perpendicular tangents to a parabola intersect on the

Options

A. directrix
B. tangent at vertex
C. axis of parabola
D. none

Answer

To solve the problem of finding the intersection point of two perpendicular tangents to the parabola \( y^2 = 4ax \), we can follow these steps:

### Step 1: Identify the points of tangency
Let the points of tangency on the parabola be \( A(T_1^2, 2aT_1) \) and \( B(T_2^2, 2aT_2) \).

### Step 2: Write the equations of the tangents
The equation of the tangent at point \( A \) is given by:
\[ y = \frac{1}{T_1}x + aT_1 \]
The equation of the tangent at point \( B \) is given by:
\[ y = \frac{1}{T_2}x + aT_2 \]

### Step 3: Set the equations equal to find the intersection point
To find the intersection point \( M \) of the two tangents, we set the two equations equal to each other:
\[ \frac{1}{T_1}x + aT_1 = \frac{1}{T_2}x + aT_2 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ \left(\frac{1}{T_1} - \frac{1}{T_2}\right)x = a(T_2 - T_1) \]

### Step 5: Solve for \( x \)
From the above equation, we can solve for \( x \):
\[ x = \frac{a(T_2 - T_1)}{\frac{1}{T_1} - \frac{1}{T_2}} = \frac{a(T_2 - T_1)T_1T_2}{T_2 - T_1} = -a \]

### Step 6: Find the corresponding \( y \) value
Substituting \( x = -a \) back into one of the tangent equations to find \( y \):
Using the tangent at \( A \):
\[ y = \frac{1}{T_1}(-a) + aT_1 = -\frac{a}{T_1} + aT_1 \]

### Step 7: Conclusion
The intersection point \( M \) will have coordinates \( (-a, y) \), which lies on the directrix of the parabola \( x = -a \).

### Final Answer
Thus, the intersection point of the two perpendicular tangents to the parabola \( y^2 = 4ax \) lies on the directrix.

---