The ordinates of the vertices of a triangle i | vedclass

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(C) The coordinates of the vertices of the triangle on the parabola $y^2 = 4ax$ are given by $\left( \frac{y_1^2}{4a}, y_1 \right)$,$\left( \frac{y_2^

Vedclass · Accepted Answer

$\frac{1}{8a}|(y_1 - y_2)(y_2 - y_3)(y_3 - y_1)|$

Vedclass · Answer

(C) The coordinates of the vertices of the triangle on the parabola $y^2 = 4ax$ are given by $\left( \frac{y_1^2}{4a}, y_1 \right)$,$\left( \frac{y_2^2}{4a}, y_2 \right)$,and $\left( \frac{y_3^2}{4a}, y_3 \right)$.The area $\Delta$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Substituting the coordinates:$\Delta = \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|$$\Delta = \frac{1}{8a} |y_1^2(y_2 - y_3) + y_2^2(y_3 - y_1) + y_3^2(y_1 - y_2)|$Using the identity $y_1^2(y_2 - y_3) + y_2^2(y_3 - y_1) + y_3^2(y_1 - y_2) = -(y_1 - y_2)(y_2 - y_3)(y_3 - y_1)$,we get:$\Delta = \frac{1}{8a} |(y_1 - y_2)(y_2 - y_3)(y_3 - y_1)|$.

The ordinates of the vertices of a triangle inscribed in the parabola $y^2 = 4ax$ are $y_1, y_2, y_3$. Find the area of the triangle.

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