Three normals are drawn to the parabola y^2=4x from the point (3,0). Find the area of the triangle formed by the feet of these three normals.

A point of the parabola is $(t^2,2t)$ and the normal there is $y=-tx+2t+t^3$. Requiring it to pass through $(3,0)$ gives $0=-3t+2t+t^3$, that is $t^3-t=0$, so $t=0,\pm1$. The feet are $(0,0)$, $(1,2)$ and $(1,-2)$. The last two lie on the vertical line $x=1$ with separation $4$, and the third vertex is at horizontal distance $1$ from that line. Area $=\tfrac12\cdot4\cdot1=2$. The shoelace formula gives the same value. Note that the feet are symmetric about the axis because the point $(3,0)$ lies on the axis, and there are three real normals because the abscissa $3$ exceeds $2$, the condition for three normals from a point $(h,0)$ on the axis of $y^2=4x$.