The tangent to an ellipse from polar coordinates

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SUMMARY

The discussion focuses on finding the tangent to an ellipse using polar coordinates, specifically with an eccentricity (ecc) of 0.60. The user initially attempts to calculate the slope of the tangent using both Cartesian and polar coordinates but encounters discrepancies in results. The correct formula for the slope of the tangent in polar coordinates is established as a combination of sine and cosine functions involving eccentricity. The user resolves the issue by clarifying that the slope is derived from dy/dx rather than dr/r*dTheta.

PREREQUISITES
  • Understanding of ellipse geometry and properties, including eccentricity.
  • Familiarity with polar coordinates and their equations.
  • Knowledge of calculus, specifically differentiation techniques.
  • Experience with Cartesian coordinates and their relationship to polar coordinates.
NEXT STEPS
  • Study the derivation of polar equations for conic sections, focusing on ellipses.
  • Learn about the relationship between Cartesian and polar coordinates in calculus.
  • Explore advanced differentiation techniques, particularly in polar coordinates.
  • Investigate applications of ellipses in physics and engineering, such as orbital mechanics.
USEFUL FOR

Mathematicians, physics students, and engineers who are working with conic sections and require a deeper understanding of tangents in polar coordinates.

Rapidrain
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I have an ellipse. Quite simple, ecc=0.60. And I'm doodling with calculus I learned 40 years ago.

I can find the tangent to the ellipse, that is, the slope of the tangent, using cartestian coordinates. At the point where the tangent skims the top of the minor axis (b) the slope is 0 and and at the semi-latus rectum slope = ecc, that is the way it should be.

Now I take the ellipse in polar coordinates around the right hand focus :

r = a*(1 - ecc**2)/(1 + ecc*cos(Theta))

to find the slope I differentiate and have :

dr/dTheta = a*ecc*(1 - ecc**2)*(sin(Theta)) / ((1 + ecc*cos(Theta))**2)

then I divide by r to get the slope of the tangent.

I am doing something wrong here because my two methods only agree at the ends of the major axis and the semi-latus rectum. At r = a, where the point is above the minor axis it has a non-zero slope.

What am I doing wrong?
 
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Got it!

dy/dx is the slope, not dr/r*dTheta,

slope is (ecc*sin(Theta)**2/(1+Ecc*cos(Theta)) + cos(Theta)) /
(ecc*sin(Theta)*cos(Theta)/(1+Ecc*cos(Theta)) - sin(Theta))
 

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