It may seem easy to compute a general formula for the distance between a point on an ellipse and a point on the xaxis (maybe it is), but I can't work it out.
Here is the data and are the restrictions.
C is an ellipse with [imath]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/imath]. On this ellipse, we have a point p (not located on the xaxis) and the normal line (perpendicular to the direction of the curve) of C at p intersects the xaxis in q. In my image, the normal line is red.
Now I want to find out what the distance between p and q is.
I have worked out a parametrisation of the ellipse: [imath]f(t)=(acos(t),bsin(t))[/imath] and that gives me [imath]f'(t)=(asin(t),bcos(t))[/imath] (this is the direction of the curve)
The direction of the normal line is then given by [imath]f''(t)=(acos(t),bsin(t))[/imath]. I get lost at that point. I have some ideas but none of those really work.
Distance between two points on/in an ellipse
Moderators: gmalivuk, Moderators General, Prelates
 Plasma_Wolf
 Posts: 93
 Joined: Mon Aug 22, 2011 8:11 pm UTC
Distance between two points on/in an ellipse
 Attachments

 Ellipse.png (6.38 KiB) Viewed 2993 times
Last edited by Plasma_Wolf on Sat Feb 18, 2012 9:01 pm UTC, edited 1 time in total.
Re: Distance between two points on/in an ellipse
Plasma_Wolf wrote:[imath]f''(t)=(a^2cos(t),b^2sin(t))[/imath]
This is not the equation of the normal. You can see this by observing that the dot product of the two lines is nonzero. Once you have the actual normal, you have
[imath](yy_0) = m(xx_0)[/imath]. You know the y_0 and x_0 from the original point, and the normal gives you m, solve for the desired intercept.
Edit: Actually, I just noticed that your parameterisation of the ellipse is incorrect as well. You may have some superfluous squares floating around.
G4!!
Grob FTW,
Hello. Smithers. You're. Quite good. At. Turning. Me. On.
Grob FTW,
Hello. Smithers. You're. Quite good. At. Turning. Me. On.
 Plasma_Wolf
 Posts: 93
 Joined: Mon Aug 22, 2011 8:11 pm UTC
Re: Distance between two points on/in an ellipse
I see what I've been doing wrong. I need a unit speed parametrization, because then I"ll get f'(t)=1, which gives me <f'(t),f'(t)>=0. Computing the derivative of that, I'll get 2*<f'(t),f''(t)>=0, which leads to f''(t) being perpendicular to the curves direction IE giving me the normal.
That gives me the question: how do I get a unit speed parametrization of an ellipse. It's easy to get this for a circle, but I've been completely unable to get a USP for an ellipse that can actually be rewritten into something easier (I want to find some common terms).
Also: I fixed the errors in my parametrizations. Having the squares there is just wrong.
That gives me the question: how do I get a unit speed parametrization of an ellipse. It's easy to get this for a circle, but I've been completely unable to get a USP for an ellipse that can actually be rewritten into something easier (I want to find some common terms).
Also: I fixed the errors in my parametrizations. Having the squares there is just wrong.
 Talith
 Proved the Goldbach Conjecture
 Posts: 848
 Joined: Sat Nov 29, 2008 1:28 am UTC
 Location: Manchester  UK
Re: Distance between two points on/in an ellipse
If you want a paramaterisation which has unit speed, you'd need to be able to have a nice formula for the arc length of an ellipse... but this is notoriously hard to calculate (see:Elliptic Integral).

 Posts: 8
 Joined: Fri Feb 10, 2012 6:07 am UTC
Re: Distance between two points on/in an ellipse
Do you know how to do implicit differentiation? It might help you here.
 Proginoskes
 Posts: 313
 Joined: Mon Nov 14, 2011 7:07 am UTC
 Location: Sitting Down
Re: Distance between two points on/in an ellipse
Plasma_Wolf wrote:It may seem easy to compute a general formula for the distance between a point on an ellipse and a point on the xaxis (maybe it is), but I can't work it out.
Here is the data and are the restrictions.
C is an ellipse with [imath]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/imath]. On this ellipse, we have a point p (not located on the xaxis) and the normal line (perpendicular to the direction of the curve) of C at p intersects the xaxis in q. In my image, the normal line is red.
Now I want to find out what the distance between p and q is.
I have worked out a parametrisation of the ellipse: [imath]f(t)=(acos(t),bsin(t))[/imath] and that gives me [imath]f'(t)=(asin(t),bcos(t))[/imath] (this is the direction of the curve)
The direction of the normal line is then given by 
I cut you off there because, as people have pointed out, your direction of the normal line is wrong.
Now you need to find a vector perpendicular to [imath]f\,'(t)[/imath]; [imath](b \cos(t), a \sin(t))[/imath] works fine. The equation for the normal line is then
[math]y  b \sin(t) = {a\over b}\tan(t) (xa \cos t).[/math]
Now substitute [imath]y=0[/imath] and solve for [imath]x[/imath]. Then [imath]q=(x,0)[/imath].
The rest is algebra and messy.
 gmalivuk
 GNU Terry Pratchett
 Posts: 25815
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: Distance between two points on/in an ellipse
The directions of the tangent and normal vectors at a particular point don't depend on the parametrization (a particle's velocity at a point will of course determine the *magnitude* of its tangent vector, but not the direction). So given the difficulty of elliptic integrals, you're better off just using the easiesttocompute parametrization
Who is online
Users browsing this forum: No registered users and 16 guests