Angle between normal vector of ellipse and the major-axis.I am trying to derive the angle made between the major or x-axis and the normal vector of an ellipse of general shape x=acos⁡(t),y=bsin⁡(t) with the parameter t reffering to Ellipse in polar coordinates. I need to solve it for any angle t. From standard reasoning I find the normal vector by its definition and checked it with the page on mathworld from wolfram and works well. Then since I know 2 points, namely a point ON the shape and a point on the normal vector I derive the angle of interest to be tan⁡(ϕ)=abtan⁡(t) Derivation. However this is very similar to the polar angle namely its simply the term a and b flipped. But when thinking about it I keep getting confused, am I correct or do I need the polar angle? If so where did I go wrong?I also found Normal to Ellipse and Angle at Major Axis but this page confused me a bit, one idea I had was they use the polar angle vs the angle I am in need of (ϕ) then I would indeed get by combing t=tan−1(abtan⁡(θ)) and ϕ=tan−1(abtan⁡(t))ϕ=tan−1(abtan⁡(tan−1big(abtan⁡θbig)))=tan−1(a2b2tan⁡θ)My excuse for my rambling, I find these angles confusing...

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Angle between normal vector of ellipse and the major-axis.I am trying to derive the angle made between the major or x-axis and the normal vector of an ellipse of general shape x=acos⁡(t),y=bsin⁡(t) with the parameter t reffering to Ellipse in polar coordinates. I need to solve it for any angle t. From standard reasoning I find the normal vector by its definition and checked it with the page on mathworld from wolfram and works well. Then since I know 2 points, namely a point ON the shape and a point on the normal vector I derive the angle of interest to be tan⁡(ϕ)=abtan⁡(t)  Derivation. However this is very similar to the polar angle namely its simply the term a and b flipped. But when thinking about it I keep getting confused, am I correct or do I need the polar angle? If so where did I go wrong?I also found Normal to Ellipse and Angle at Major Axis but this page confused me a bit, one idea I had was they use the polar angle vs the angle I am in need of (ϕ) then I would indeed get by combing t=tan−1(abtan⁡(θ)) and ϕ=tan−1(abtan⁡(t))ϕ=tan−1(abtan⁡(tan−1big(abtan⁡θbig)))=tan−1(a2b2tan⁡θ)My excuse for my rambling, I find these angles confusing...

glennastlf7 · Accepted Answer

The tangent vector to the ellipse at parameter value t is (−asin⁡t,bcos⁡t), so the tangent of the angle ψ it makes with the positive x-axis is given bytan⁡ψ=−bacot⁡t.The angle ϕ to the outward normal vector is ψ−π2, sotan⁡ϕ=tan⁡(ψ−π2)=−cot⁡(ψ)=abtan⁡t,as desired.

Angle between normal vector of ellipse and the

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