/// <summary>
/// Returns all the perpendicular points on the ellipse from a given point 'from'
/// </summary>
public Point2d[] Perpendicular(Point2d from)
{
// Solved by Robert.P. in december 2012
// Note on solutions:
// Quartic coefficients gotten from applying lagrange multiplier to minimize (x-i)^2+(y-j)^2
// with x^2/a^2+y^2/b^2-1=0 as constraint (a=1 because we work in standard position).
// This gives a system of three equations F_x,F_y,F_lambda, which were solved with
// resultant theory using 'eliminate' in maxima
//work in standard position, retranslate solutions last
Transform2d tostd = ToStandardPosition;
from = from.GetTransformed(tostd);
double b = sigratio, b2 = b * b, b4 = b2 * b2;
double i = from.X, i2 = i * i;
double j = from.Y, j2 = j * j;
double x4 = b4 - 2 * b2 + 1;
double x3 = 2 * b2 * i - 2 * i;
double x2 = b2 * j2 + i2 - b4 + 2 * b2 - 1;
double x1 = 2 * i - 2 * b2 * i;
double x0 = -i2;
double[] sols = RealPolynomial.SolveQuartic2(x4, x3, x2, x1, x0, 1e-16);
if (sols == null)
{
return(null);
}
Point2dSet respts = new Point2dSet();
foreach (double x in sols)
{
double y = (1 - x * x) * b2;
if (y < 0.0)
{
continue;
}
y = Math.Sqrt(y);
for (int l = 0; l < 2; l++)
{
//both posetive and negative y:s can be solutions. Check with each possible
//point that its perpendicular to ellipse (subtracting the inverse ellipse slope (=normal slope) with the slope from 'from' point)
double err;
err = y * (from.X - x) - x * b2 * (from.Y - y);
if (Math.Abs(err) < 1e-6)
{
respts.Add(new Point2d(x, y));
}
y = -y; //test negative solution as well
}
}
respts.Transform(tostd.Inversed);
return(respts.ToArray());
}
