Distance between two points on/in an ellipse

[Plasma_Wolf]

It may seem easy to compute a general formula for the distance between a point on an ellipse and a point on the x-axis (maybe it is), but I can't work it out.

Here is the data and are the restrictions.
C is an ellipse with \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. On this ellipse, we have a point p (not located on the x-axis) and the normal line (perpendicular to the direction of the curve) of C at p intersects the x-axis 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: f(t)=(acos(t),bsin(t)) and that gives me f'(t)=(-asin(t),bcos(t)) (this is the direction of the curve)
The direction of the normal line is then given by f''(t)=(-acos(t),-bsin(t)). I get lost at that point. I have some ideas but none of those really work.

[OverBored]

f''(t)=(-a^2cos(t),-b^2sin(t))

This is not the equation of the normal. You can see this by observing that the dot product of the two lines is non-zero. Once you have the actual normal, you have
(y-y_0) = m(x-x_0). 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.

[Plasma_Wolf]

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.

[Talith]

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).

[SeaCalMaster]

Do you know how to do implicit differentiation? It might help you here.

[Proginoskes]

It may seem easy to compute a general formula for the distance between a point on an ellipse and a point on the x-axis (maybe it is), but I can't work it out.

Here is the data and are the restrictions.
C is an ellipse with \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. On this ellipse, we have a point p (not located on the x-axis) and the normal line (perpendicular to the direction of the curve) of C at p intersects the x-axis 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: f(t)=(acos(t),bsin(t)) and that gives me f'(t)=(-asin(t),bcos(t)) (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 f\,'(t); (b \cos(t), a \sin(t)) works fine. The equation for the normal line is then

y - b \sin(t) = {a\over b}\tan(t) (x-a \cos t).

Now substitute y=0 and solve for x. Then q=(x,0).

The rest is algebra and messy.

[gmalivuk]

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 easiest-to-compute parametrization