If `t_1, t_2` and `t_3` are distinct, then the points `(t_1, 2at_1 + at_1^3), (t_2, 2at_2 + at_2^3)` and `(t_3, 2at_3 + at_3^3)` are collinear, when

Options

A. `t_1 t_2t_3 = 1`
B. `t_1 + t_2 + t_3 = t_1 t_2 t_3`
C. `t_1 + t_2 + t_3 = 0`
D. `t_1 + t_2 + t_3 = 1`

Answer

To determine the condition under which the points \((t_1, 2at_1 + at_1^3)\), \((t_2, 2at_2 + at_2^3)\), and \((t_3, 2at_3 + at_3^3)\) are collinear, we can follow these steps:

### Step 1: Understand the concept of collinearity
Three points are collinear if the slope between any two pairs of points is the same. Thus, we need to find the slopes between the points and set them equal.

### Step 2: Calculate the slope of line segment AB
Let \(A = (t_1, 2at_1 + at_1^3)\) and \(B = (t_2, 2at_2 + at_2^3)\). The slope \(m_{AB}\) between points A and B is given by:
\[ m_{AB} = \frac{(2at_2 + at_2^3) - (2at_1 + at_1^3)}{t_2 - t_1} \]
This simplifies to:
\[ m_{AB} = \frac{2a(t_2 - t_1) + a(t_2^3 - t_1^3)}{t_2 - t_1} \]
Using the difference of cubes formula, \(t_2^3 - t_1^3 = (t_2 - t_1)(t_2^2 + t_1t_2 + t_1^2)\), we can rewrite the slope as:
\[ m_{AB} = 2a + a(t_2^2 + t_1t_2 + t_1^2) \]

### Step 3: Calculate the slope of line segment BC
Now, let \(C = (t_3, 2at_3 + at_3^3)\). The slope \(m_{BC}\) between points B and C is given by:
\[ m_{BC} = \frac{(2at_3 + at_3^3) - (2at_2 + at_2^3)}{t_3 - t_2} \]
This simplifies to:
\[ m_{BC} = \frac{2a(t_3 - t_2) + a(t_3^3 - t_2^3)}{t_3 - t_2} \]
Again, using the difference of cubes:
\[ m_{BC} = 2a + a(t_3^2 + t_2t_3 + t_2^2) \]

### Step 4: Set the slopes equal
For the points to be collinear, we need:
\[ m_{AB} = m_{BC} \]
This gives us the equation:
\[ 2a + a(t_2^2 + t_1t_2 + t_1^2) = 2a + a(t_3^2 + t_2t_3 + t_2^2) \]
Subtracting \(2a\) from both sides and simplifying:
\[ t_2^2 + t_1t_2 + t_1^2 = t_3^2 + t_2t_3 + t_2^2 \]
This reduces to:
\[ t_2^2 + t_1t_2 + t_1^2 - t_2^2 = t_3^2 + t_2t_3 \]
Thus:
\[ t_1t_2 + t_1^2 = t_3^2 + t_2t_3 \]

### Step 5: Rearranging the equation
Rearranging gives us:
\[ t_1t_2 - t_2t_3 = t_3^2 - t_1^2 \]
Factoring both sides leads to:
\[ t_2(t_1 - t_3) = (t_3 - t_1)(t_3 + t_1) \]

### Step 6: Conclusion
The points are collinear if:
\[ t_2(t_1 - t_3) = (t_3 - t_1)(t_3 + t_1) \]