# Find the area of triangle inscribed in the parabola $y^2=4ax$ with vertices $(x_1,y_1),(x_2,y_2)$ & $(x_3,y_3)$?

$\begin{array}{1 1}(a)\;\large\frac{1}{8a}\normalsize \mid(y_1-y_2)(y_2-y_3)(y_3-y_1)\mid\\(b)\;\large\frac{1}{8a}\normalsize \mid(y_1+y_2)(y_2+y_3)\mid\\(c)\;\large\frac{1}{8a}\normalsize \mid(y_1-y_2)(y_2-y_3)\mid\\(d)\;\large\frac{1}{2a}\normalsize \mid(y_1-y_2)(y_2-y_3)\mid\end{array}$

$(x_1,y_1),(x_2,y_2)$ & $(x_3,y_3)$ are the vertices which lie on the parabola hence
$x_1=\large\frac{y_1^2}{4a}$
$x_2=\large\frac{y_2^2}{4a}$
$x_3=\large\frac{y_3^2}{4a}$
Now area of triangle with $(\large\frac{y_1^2}{4a}$$,y_1),(\large\frac{y_2^2}{4a}$$,y_2),(\large\frac{y_3^2}{4a}$$,y_3) is \large\frac{1}{2}$$\begin{vmatrix}\large\frac{y_1^2}{4a}&y_1&1\\\large\frac{y_2^2}{4a}&y_2&1\\\large\frac{y_3^2}{4a}&y_3&1\end{vmatrix}$
Hence area of triangle=$\large\frac{1}{2}\mid\frac{y_1^2}{4a}$$(y_2-y_3)-y_1(\large\frac{y_2^2}{4a}-\frac{y_3^2}{4a})$$+1(\large\frac{y_2^2y_3}{4a}-\frac{y_2y_3^2}{4a})\mid$
$\Rightarrow \large\frac{1}{2}\mid \big(\large\frac{y_1^2y_2-y_1^2y_3-y_1y_2^2+y_1y_3^2+y_2^2y_3-y_2y_3^2}{4a}\big)\mid$
$\Rightarrow \large\frac{1}{8a}$$\mid(y_1-y_2)(y_2-y_3)(y_3-y_1)\mid$
Hence (a) is the correct answer.