Q.Prove that the area of the triangle inscribed in the parabola $y^2 = 4ax$ is $\dfrac{1}{8a} |(y_1 - y_2)(y_2 - y_3)(y_3 - y_1)|$ sq. units where $y_1, y_2, y_3$ are the ordinates of its vertices.

Write each vertex as $\left(\dfrac{y_i^2}{4a},y_i\right)$ on the parabola, plug into the standard triangle-area formula, and factor the resulting expression using the identity $a^2(b-c)+b^2(c-a)+c^2(a-b)=-(a-b)(b-c)(c-a)$.

A point on $y^2=4ax$ with ordinate $y_i$ has abscissa $x_i=\dfrac{y_i^2}{4a}$, so the three vertices are $\left(\dfrac{y_1^2}{4a},y_1\right),\left(\dfrac{y_2^2}{4a},y_2\right),\left(\dfrac{y_3^2}{4a},y_3\right)$.

The area of a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is

$\text{Area}=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$

Substituting the $x_i$:

$\text{Area}=\frac12\left|\frac{y_1^2}{4a}(y_2-y_3)+\frac{y_2^2}{4a}(y_3-y_1)+\frac{y_3^2}{4a}(y_1-y_2)\right|=\frac{1}{8a}\left|y_1^2(y_2-y_3)+y_2^2(y_3-y_1)+y_3^2(y_1-y_2)\right|.$

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