Question

If $y_1,y_2$ and $y_3$ are the ordinates of the vertices of a triangle inscribed in the parabola $y^2=4ax$ then its area is

• A. $\frac{1}{2a}|(y_1-y_2)(y_2-y_3)(y_3-y_1)|$
• B. $\frac{1}{4a}|(y_1-y_2)(y_2-y_3)(y_3-y_1)|$
• C. $\frac{1}{8a}|(y_1-y_2)(y_2-y_3)(y_3-y_1)|$
• D. none

Answer 1

The correct option is C $\frac{1}{8a}|(y_1-y_2)(y_2-y_3)(y_3-y_1)|$

Vertices of the triangle are $(\frac{y_1^2}{4a},y_1)(\frac{y_2^2}{4a},y_2)\&(\frac{y_3^2}{4a},y_3)$
Area of the triangle = $\frac{1}{2}\left|\begin{array}{c}\frac{1}{4a}(y_1^2-y_2^2)y_1-y_2\\ \frac{1}{4a}(y_1^2-y_2^2)y_1-y_3\end{array}\right|=\frac{1}{2}\left(\frac{1}{4a}\right)|(y_1^2-y_2^2)(y_1-y_3)|$
= $\frac{1}{8a}|(y_1-y_2)(y_2-y_3)(y_3-y_1)|$