How to find the 4 angles from center of a rotated ellipse to it 4 extermaties

Question

I am trying to find an algorithm to derive the 4 angles, from the centre of a rotated ellipse to its extremities. In other words, from the centre to the points where the bounding box touches the ellipse's line.

I have already figured out how to get the bounding box by using:

leftX/rightX  =  h± sqrt(a*a*cos(PI)*cos(PI) + b*b*sin(PI)*sin(PI))
topY/bottomY  =  k± sqrt(a*a*sin(PI)*sin(PI) + b*b*cos(PI)*cos(PI))

The above gives me the bounding box AND:

the x of the left and right points

the y of the top and bottom points

But I need the x and y of the left and right points and the x and y of the top and bottom points in order to calculate the angles.

I feel like I missing something simple but could not find it.

The Image illustrates the point at the top of the bounding box.

Answer 1

Let ellipse semiaxes are a, b , center point is (cx, cy) and rotation angle is fi. Then coordinates on ellipse circumference at parameter t are:

 x = a * Cos(t) * Cos(fi) - b * Sin(t) * Sin(fi) + cx
 y = a * Cos(t) * Sin(fi) + b * Sin(t) * Cos(fi) + cy

To get extremal points, we can differentiate coordinates by t and set derivatives to zero

 a * Sin(t) * Cos(fi) = - b * Cos(t) * Sin(fi)   //  dx/dt
 a * Sin(t) * Sin(fi)  =  b * Cos(t) * Cos(fi) //  dy/dt

 tg(t) = -b/a * Tg(fi)
 t(l,r)  = Pi * k + Atan(-b/a * Ttg(fi)  //left and right, k=0,1

 tg(t) = b/a * Ctg(fi)
 t(t,b) = Pi * k + Atan(b/a * Ctg(fi))  //top and bottom, k = 0,1

 ytop = a * Cos(t(t)) * Sin(fi) + b * Sin(t(t)) * Cos(fi) + cy
 and so on

real example for a: 200; b: 100; fi: Pi/6 CW; generated by quick-made Delphi code