The area of the triangle formed by the tangent to the parabola $y^2=4ax$ at the point $(a{t}^2,2at)$ and the coordinate axes is:

Solution

1. The equation of the tangent to the parabola $y^2=4ax$ at the point $(a{t}^2,2at)$ is given by $ty=x+a{t}^2$.
2. Setting $y=0$ gives the x-intercept $x=-a{t}^2$, and setting $x=0$ gives the y-intercept $y=at$.
3. The area of the triangle formed by these intercepts and the origin is $\frac{1}{2}|(-a{t}^2)(at)|=\frac{1}{2}{a}^2|t^3|$.
4. Assuming $t>0$, the area simplifies to $\frac{1}{2}{a}^2{t}^3$, which corresponds to the magnitude of the geometric area.
5. Hence the answer is (A).