Monday, June 27, 2011

Rotating a hyperbola: general case

Just to finish up quickly with the analysis of hyperbolas from the other day (here and here), following Stewart (pdf), we derived the following equations for x and y in terms of a coordinate system based on u and v (orthogonal, rotated through angle θ):

 `x = u cosθ - v sinθy = u sinθ + v cosθ`

The general formula for a quadratic is:

 `Ax2 + Bxy + Cy2 + Dx + Ey + F = 0`

When we transform to the new coordinates, we get:

 `Ax2 = A[u2 cos2θ - 2uv sinθ cosθ + v2 sin2θ]Bxy = B[u2 cosθ sinθ + uv cos2θ - uv sin2θ - v2 sinθ cosθCy2 = C[u2 sin2θ + 2uv sinθ cosθ + v2 cos2θ`

Gathering the terms in uv we obtain the coefficients:

 `(C-A)(2sinθ cosθ) + B(cos2θ - sin2θ)`

Remember the double angle formulas:

 `sin(s+t) = sin s cos t + cos s sin tsin(2θ) = 2 sinθ cosθcos(s+t = cos s cos t - sin s sin tcos(2θ) = cos2θ - sin2θ`

So we obtain:

 `(C-A) sin(2θ) + B cos(2θ)`

for the coefficients of xy. These must equal zero for all the xy terms to disappear. Thus:

 `tan(2θ) = B / (C-A)`

This approach runs into a problem if C = A, as it does for our example:

 `xy = 1`

But we can just invert the step at the end:

 `cot(2θ) = (C-A)/B = 0`

The cotangent is zero when the cosine is zero, e.g. for

 `2θ = π/2`

Thus, if

 `θ = π/4`

then all the xy terms vanish, as we found before.