once I have the slope of the tangent line I simply take the negative reciprocal (perpendicular) of that slope and I have my normal.

Don't you have to integrate it back once you're done? And include a constant offset after that. I'm a bit rusty with this, but here's my worksheet:

> x

^{2}/a

^{2}+ y

^{2}/b

^{2}- 1 = 0

> 2x/a

^{2}+ (2y/b

^{2}) * (dy/dx) = 0 // first differential

> (dy/dx) = - (x * b

^{2}) / (y * a

^{2}) // slope

> (dy/dx) = (y * a

^{2}) / (x * b

^{2}) // normal in differential form

> (dy/dx) = (pt.y * a

^{2}) / (pt.x * b

^{2}) // Now replace with actual intersection coordinates

> (dy/dx) = k // The RHS is going to be a constant value now, say, k

> y = k * x + C // integrate back to get actual line

> C = pt.y - pt.x * k // since the intersection point also lies on the line

Hope this helps

Cheers

~dd~