How to bring the tangent of a parametric function on a plane to a vector form? How to find the normal vector at a point on a flat graph?

Question

Mathematics

The task is to find the direction vector for the tangent to a point on the graph of a parametric function on a plane.
In the classical form, the tangent is considered through the derivative, and in the answer we get the tangent of the slope of the tangent angle relative to the X axis. From here it immediately comes out minus that at the angles of 90 ° and 270 ° the tangent is not defined. Secondly, it is not clear how to normalize the resulting value. For the second day, I have been puzzling over how to express this tangent through a single vector, while being as compact and resource efficient as possible, since this code will be used in the vertex shader.
The end goal is to get a normal to a given point, now I do it from a tangent by rotating (-1/f'(x)). But it is necessary to somehow express this normal in vector form. If there is a direct way to calculate the normal vector at a point on a graph, I'd love to hear it!

Answer 1

If the curve is given in the parametric form X=X(t), Y=Y(t), then everything is simpler. The tangent is (dY/dt)/(dX/dt), and the normal vector is (dX/dt)/(dY/dt).