Show that the area formed by the normals to `y^2=4ax` at the points `t_1,t_2,t_3` is

Answer

To solve the problem of finding the area formed by the normals to the parabola \( y^2 = 4ax \) at the points \( t_1, t_2, t_3 \), we can follow these steps:

### Step 1: Determine the points on the parabola
The points on the parabola corresponding to the parameters \( t_1, t_2, t_3 \) are given by:
\[ P(t_i) = (at_i^2, 2at_i) \quad \text{for } i = 1, 2, 3 \]

### Step 2: Find the equations of the normals
The equation of the normal to the parabola at a point \( P(t_i) \) is given by:
\[ y - 2at_i = -\frac{1}{2a} (x - at_i^2) \]
Rearranging this gives:
\[ y = -\frac{1}{2a}x + \left(2at_i + \frac{at_i^2}{2a}\right) = -\frac{1}{2a}x + at_i + 2at_i \]
Thus, the equation of the normal can be simplified to:
\[ y = -\frac{1}{2a}x + 3at_i \]

### Step 3: Find the intersection points of the normals
We need to find the intersection points of the normals at \( t_1, t_2, t_3 \):
1. For \( t_1 \): \( y = -\frac{1}{2a}x + 3at_1 \)
2. For \( t_2 \): \( y = -\frac{1}{2a}x + 3at_2 \)
3. For \( t_3 \): \( y = -\frac{1}{2a}x + 3at_3 \)

### Step 4: Set up the vertices of the triangle
The vertices of the triangle formed by these normals can be defined as:
- Vertex A: \( \left(2a + at_1^2, 3at_1\right) \)
- Vertex B: \( \left(2a + at_2^2, 3at_2\right) \)
- Vertex C: \( \left(2a + at_3^2, 3at_3\right) \)

### Step 5: Use the determinant formula to find the area
The area \( A \) of the triangle formed by the points \( A, B, C \) can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \]
Substituting the coordinates:
\[ A = \frac{1}{2} \left| \begin{vmatrix} 2a + at_1^2 & 3at_1 & 1 \\ 2a + at_2^2 & 3at_2 & 1 \\ 2a + at_3^2 & 3at_3 & 1 \end{vmatrix} \right| \]

### Step 6: Calculate the determinant
Calculating the determinant:
\[ = \left(2a + at_1^2\right)\left(3at_2 - 3at_3\right) - \left(3at_1\right)\left(2a + at_2^2 - 2a - at_3^2\right) + 1\left(3at_1(2a + at_3^2) - 3at_2(2a + at_1^2)\right) \]
This simplifies to:
\[ = \frac{a^2}{2} \left| (t_1 - t_2)(t_2 - t_3)(t_3 - t_1) \right| \cdot (t_1 + t_2 + t_3) \]

### Final Result
Thus, the area of the triangle formed by the normals to the parabola at the points \( t_1, t_2, t_3 \) is:
\[ A = \frac{a^2}{2} |(t_1 - t_2)(t_2 - t_3)(t_3 - t_1)| \cdot (t_1 + t_2 + t_3) \]