How do I find the normal and tangent equations? x^2/4 + (y+2)^2 = 1 (ellipse) "Find the parametrized equation to the tangent and normal in point (-2, -2)." I believe the tangent is [-2, -2] + t[0, -1] (although I am not sure), and I believe the normal is [-2, -2] + t[-1, 0] (How do you get to the solution? I just "saw" it, but I cant figure out how to mathematically do it!) Tried N = N(vector) / N(scalar), but .. nope

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How do I find the normal and tangent equations? x^2/4 + (y+2)^2 = 1 (ellipse) "Find the parametrized equation to the tangent and normal in point (-2, -2)." I believe the tangent is [-2, -2] + t[0, -1] (although I am not sure), and I believe the normal is [-2, -2] + t[-1, 0] (How do you get to the solution? I just "saw" it, but I cant figure out how to mathematically do it!) Tried N = N(vector) / N(scalar), but .. nope

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Have you learnt differentiation yet?
You can try the polar line of the elipse, then consider the family of lines on the desired point with the conditions given.
My notes; http://i.imgur.com/czFgEuq.png (finding tangent) http://i.imgur.com/rBAISmO.png (finding normal)

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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?
for two dimensions, yes. for higher dimensions you have to use that formula N = T ' (t) / | T ' (t) |
remember from algebra to find the slope of a perpendicular line, we use inverse reciprocal of the given line's slope
I need to polish on old math (been forever since using them). I had trouble finding examples for two dimensions
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
I understand - I have found tons of material on "planes" - not what I wanted for this task , had to find the "simple" 2D plane normals, I just dont remember much
so t is also underneath - divided?
\[[-2, -2] - \frac{ 1 }{ t*[0, -1] }\] ? Am I understanding it right?
yes that is correct

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