Question

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

Answer 1

Have you learnt differentiation yet?

Answer 2

You can try the polar line of the elipse, then consider the family of lines on the desired point with the conditions given.

Answer 3

My notes;
http://i.imgur.com/czFgEuq.png (finding tangent)
http://i.imgur.com/rBAISmO.png (finding normal)

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

r'(-2,-2) is [0, -1]

Where can I find the formula/explaination for the b part you suggested, jazzdd?

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

I think I already wrote that in the notes

Answer 12

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

Answer 13

the slope of the normal is perpendicular to the tangent line

Answer 14

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

Answer 15

for two dimensions, yes. for higher dimensions you have to use that formula
N = T ' (t) / | T ' (t) |

Answer 16

remember from algebra
to find the slope of a perpendicular line, we use inverse reciprocal of the given line's slope

Answer 17

I need to polish on old math (been forever since using them).
I had trouble finding examples for two dimensions

Answer 18

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

Answer 19

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

Answer 20

yes

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

Answer 21

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

Answer 22

so t is also underneath - divided?

Answer 23

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

Answer 24

yes that is correct