0

There is a point (x,y) in the elliptical orbit inclined at theta angle. I know the x, y coordinates and the center coordinates and Major Minor. In this case, is there an equation to find the tangent line?

Amour Math
  • 19
  • 6
  • If you have angle of inclination, center and axes you can write the equation of the ellipse. Once you have the equation, use implicit differentiation to find the slope of a tangent line. Finally, find equation of a tangent that goes thru a particular point. – Vasili Jun 08 '23 at 12:56
  • (h,k) = Center and theta=17 ● a=400/2 ● b=314/2 ● h=500 ● k=500 ● x=308.739049 ● y= 441.525659 ● ((x-h)Cos(theta)+(y-k)Sin(theta))^2/a^2 + (-(x-h)Sin(theta)+(y-k)Cos(theta))^2/b^2=1 ● Q=(1/a)((x-h)Cos(theta)+(y-k)Sin(theta)) ● R=(1/b)(-(x-h)Sin(theta)+(y-k)Cos(theta)) ● x=aQCos(theta)-bRSin(theta)+h = 308.739049 ● y=aQSin(theta)+bRCos(theta)+k = 441.525659 ● Tangent_Angle=ATan( bQ / (-aR) )+theta = -73 – Amour Math Jun 08 '23 at 19:49

1 Answers1

1

If $S$ is the center, $\theta$ the angle of inclination and $a,b$ the semi-axes lengths, the equation would be $$\frac{((x-x(S)) \cos{\theta}+(y-y(S)) \sin{\theta})^2}{a^2}+\frac{(-(x-x(S)) \sin{\theta}+(y-y(S))\cos{\theta})^2}{b^2}=1,$$ and the tangent at $(h,k)$ is given from the expanded form $Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$ by the usual $Axh+B(xk+yh)+Cyk+D(x+h)+E(y+k)+F=0.$

Jan-Magnus Økland
  • 5,149
  • Thank you from the bottom of my heart. I searched the websites of various countries around the world and found this way. (h,k) = Center and theta=17 ● a=400/2 ● b=314/2 ● h=500 ● k=500 ● x=308.739049 ● y= 441.525659 ● ((x-h)Cos(theta)+(y-k)Sin(theta))^2/a^2 + (-(x-h)Sin(theta)+(y-k)Cos(theta))^2/b^2=1 ● Q=(1/a)((x-h)Cos(theta)+(y-k)Sin(theta)) ● R=(1/b)(-(x-h)Sin(theta)+(y-k)Cos(theta)) ● x=aQCos(theta)-bRSin(theta)+h = 308.739049 ● y=aQSin(theta)+bRCos(theta)+k = 441.525659 ● Tangent_Angle=ATan( bQ / (-aR) )+theta = -73 – Amour Math Jun 08 '23 at 19:55
  • @AmourMath The above method in geogebra gives the same result. – Jan-Magnus Økland Jun 09 '23 at 08:12
  • Oh, yep : ) Thank you for showing me the way : ) – Amour Math Jun 11 '23 at 07:22