If `y_(1),y_(2),andy_(3)` are the ordinates of the vertices of a triangle inscribed in the parabola `y^(2)=4ax`, then its area is

Options

A. `(1)/(2a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`
B. `(1)/(4a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`
C. `(1)/(8a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`
D. none of these

Explanation

To find the area of the triangle inscribed in the parabola \( y^2 = 4ax \) with vertices having ordinates \( y_1, y_2, \) and \( y_3 \), we can follow these steps:

### Step 1: Identify the coordinates of the triangle vertices
The vertices of the triangle can be represented as:
- \( A( \frac{y_1^2}{4a}, y_1) \)
- \( B( \frac{y_2^2}{4a}, y_2) \)
- \( C( \frac{y_3^2}{4a}, y_3) \)

### Step 2: Use the formula for the area of a triangle
The area \( A \) of a triangle given its vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 3: Substitute the coordinates into the area formula
Substituting the coordinates of points \( A, B, \) and \( C \):
\[ A = \frac{1}{2} \left| \frac{y_1^2}{4a}(y_2 - y_3) + \frac{y_2^2}{4a}(y_3 - y_1) + \frac{y_3^2}{4a}(y_1 - y_2) \right| \]

### Step 4: Factor out common terms
Factor out \( \frac{1}{4a} \):
\[ A = \frac{1}{8a} \left| y_1^2(y_2 - y_3) + y_2^2(y_3 - y_1) + y_3^2(y_1 - y_2) \right| \]

### Step 5: Simplify the expression
Now, we can simplify the expression inside the absolute value:
\[ A = \frac{1}{8a} \left| y_1^2(y_2 - y_3) + y_2^2(y_3 - y_1) + y_3^2(y_1 - y_2) \right| \]

### Final Result
Thus, the area of the triangle inscribed in the parabola \( y^2 = 4ax \) with ordinates \( y_1, y_2, y_3 \) is:
\[ A = \frac{1}{8a} \left| y_1^2(y_2 - y_3) + y_2^2(y_3 - y_1) + y_3^2(y_1 - y_2) \right| \]