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Have you learnt differentiation yet?

first find the parametric equations for the ellipse
x = h + a cos t
y = k + b sin t

Already got the parametric equation for the equation;
http://i.imgur.com/73ELALw.png

wouldn't it be
part a) (-2, -2) + t * r ' (-2,-2)
part b) (-2, -2) + t * -1 / r ' (-2,-2)

r'(-2,-2) is [0, -1]
Where can I find the formula/explaination for the b part you suggested, jazzdd?

nevermind, we have to find t such that it comes to that point.

to get to the point (-2,-2) we need t = π

part a) (-2, -2) + t * r ' (π)
part b) (-2, -2) + t * -1 / r ' (π)

I think I already wrote that in the notes

But I do not understand the part b) formula - where is it from, how did you conclude it

the slope of the normal is perpendicular to the tangent line

So I just multiply by -1 and divide by r'(..) to get any normal?

How would you write part b as an answer?
\[r(t_0)-t*r'(t_0)^{-1} ?\] (t_0 being pi)

this is only valid in 2 dimensions
finding a normal in 3 dimension is more involved

yes
r(t_0) + -1/ r ' (t _0) * t

so t is also underneath - divided?

\[[-2, -2] - \frac{ 1 }{ t*[0, -1] }\] ? Am I understanding it right?

yes that is correct