
Re: Ellipse and line intersection points
Posted:
Jun 25, 2014 7:09 PM


In article <lofijs$64g$1@dontemail.me>, Cristiano <cristiapi@NSgmail.com> wrote:
> .... > You answered as if it were a homework, but I'm not a student (and I'm > not a mathematician)....
Not at all. It's much too messy to be any reasonable homework.
> > > Solve the last pair of equations for x(E) and y(E) in terms of xr and yr. > > > > Solve the first pair of equations for cos(E) and sin(E) in terms of x(E) > > and y(E) and hence in terms of xr and yr. > > cos(E)= ((xr  k2 * y(E)) / k1 + ae) / a > > y(E)= (yr  k3/k1 * (xr  k2 * y(E))) / k4 > > I don't know how to isolate y(E). > > Cristiano
It might be easier to solve the system in each numerical case you want. But if you really want the general formula, prepare for it to be very complicated.
Solve your last pair of linear equations for x(E) and y(E) in terms of xr and yr:
x(E) = (k4*xr  k2*yr)/(k1*k4  k2*k3)
y(E) = (k3*xr + k1*yr)/(k1*k4  k2*k3)
Similarly solve the first pair for cos(E) and sin(E) in terms of x(E) and y(E). Substitute the above solutions into the new ones to get cos(E) and sin(E) in terms of xr and yr (messy!). Then use (cos(E))^2 + (sin(E))^2 = 1 to get a big equation in xr and yr (messier still). Then put xr = k to get a quadratic equation for yr, and solve that. (even more messy).
The result will be a very large formula with a square root in it. I'm avoiding the details, not because I think it's homework, but because it would take more time and trouble than I want to spend on it. Sorry.
Ken Pledger.

