Math Is Fun Forum

hi Bob,

I had to sort things out first, but now I have the rotation working:

You start with the hyperbola x.x / a.a - y.y / b.b - 1 = 0  [ x.x means x squared etc.]

This is a hyperbola with its center in the origin (0,0)

It's general equation is: A.x.x + C.y.y + F = 0

thus:
A = 1/a.a
C = -1/b.b
F = -1
(B = D = E = 0)

Rotation with angle theta means replacing
   x by  x.cos(theta) + y.sin(theta)   and
   y by  y.cos(theta) -  x.sin(theta)

Doing the algebra you find new coëfficients that describe the rotated hyperbola.

In Java:

// this is a method in my Conics-class, using global variables A thru F, that I adjust, forming a new Conics 

public void     rotate(double theta) {
   
      double c = cos(theta);
      double s = sin(theta);
     
      A1 = A*c*c + C*s*s;
      B1 = 2.0*c*s*(A-C);
      C1 = A*s*s + C*c*c;
   
      if (B != 0) {
         B1 = B1 + B*(c*c-s*s);
         A1 = A1 - B*c*s;
         C1 = C1 + B*c*s;
      }
      new Conics(A1, B1, C1, 0, 0, -1); // the class that produces a Shape (where A=A1, B=B1 andsoforth)
   }
when the original hyperbola already has a nonzero B-coefficient (was already rotated by some angle), you have to solve

B1 = B(xc+ys)(yc-xs) [same replacement of x and y as above]
 
this produces extra x.x and y.y terms, that you must add to A and C respectively.
   
success.

peter