The area of the triangle formed by the tangents at the points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ on the parabola $y^2=4ax$ is:

Solution

1. Let the points on the parabola $y^2=4ax$ be $(a{t}_i^2,2a{t}_i)$ for $i=1,2,3$, where $y_i=2a{t}_i$.
2. The intersection point of the tangents at $t_i$ and $t_j$ is given by $(a{t}_i{t}_j,a(t_i+t_j))$.
3. The area of the triangle with vertices $(a{t}_1{t}_2,a(t_1+t_2))$, $(a{t}_2{t}_3,a(t_2+t_3))$, and $(a{t}_3{t}_1,a(t_3+t_1))$ is $\frac{1}{2}|a^2(t_1-t_2)(t_2-t_3)(t_3-t_1)|$.
4. Substituting $t_i=\frac{y_i}{2a}$ into the area formula yields $\frac{1}{2}|a^2(\frac{y_1-y_2}{2a})(\frac{y_2-y_3}{2a})(\frac{y_3-y_1}{2a})|=\frac{1}{8a}|(y_1-y_2)(y_2-y_3)(y_3-y_1)|$.
5. Hence the answer is (A).