Find the area of that triangle whose vertices are `(at_(1)^(2),2at_(1)),(at_(2)^(2),2at_(2))and(at_(3)^(2),2at_(3)).`

Answer

To find the area of the triangle with vertices given by the coordinates \((at_1^2, 2at_1)\), \((at_2^2, 2at_2)\), and \((at_3^2, 2at_3)\), we can use the formula for the area of a triangle formed by three points in the coordinate plane:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 1: Identify the coordinates
Let:
- \( (x_1, y_1) = (at_1^2, 2at_1) \)
- \( (x_2, y_2) = (at_2^2, 2at_2) \)
- \( (x_3, y_3) = (at_3^2, 2at_3) \)

### Step 2: Substitute the coordinates into the area formula
Substituting the coordinates into the area formula gives us:

\[ \text{Area} = \frac{1}{2} \left| at_1^2(2at_2 - 2at_3) + at_2^2(2at_3 - 2at_1) + at_3^2(2at_1 - 2at_2) \right| \]

### Step 3: Simplify the expression
Factoring out common terms:

\[ = \frac{1}{2} \left| 2a \left( at_1^2(t_2 - t_3) + at_2^2(t_3 - t_1) + at_3^2(t_1 - t_2) \right) \right| \]

This simplifies to:

\[ = a \left| at_1^2(t_2 - t_3) + at_2^2(t_3 - t_1) + at_3^2(t_1 - t_2) \right| \]

### Step 4: Factor out \(a\)
We can factor out \(a\) from the absolute value:

\[ = a^2 \left| t_1^2(t_2 - t_3) + t_2^2(t_3 - t_1) + t_3^2(t_1 - t_2) \right| \]

### Step 5: Final expression for the area
Thus, the area of the triangle is given by:

\[ \text{Area} = a^2 \left| t_1^2(t_2 - t_3) + t_2^2(t_3 - t_1) + t_3^2(t_1 - t_2) \right| \]