I'm given a periodic parametric curve $P = ( x(t),y(t) )$, where $t \in [0, 2\pi)$ and $P(2\pi)=P(0)$. I have a point $F = (x,y)$ that is not on that surface. Could someone tell me how to find the point $B = (x_p,y_p)$ on that surface that is normal to F? Is there a closed form expression I could use?
Note: I have the same problem in 3D, with a surface $P = ( x(u,v),y(u,v),z(u,v) )$, where $u \in [0, 2\pi)$ and $v \in [0, \pi)$.
Thank you for your time!