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Angle to remove xy term from ax²+2hxy+by²=0

Short answer

Rotate by θ such that tan2θ=2h/(a−b). If a=b, θ=45°. The resulting equation is λ₁X²+λ₂Y²=0 with λ₁,λ₂ = ½[(a+b)±√((a−b)²+4h²)].

Find the angle through which the axes must be turned so that the equation ax² + 2hxy + by² = 0 may become an equation having no term involving xy.

Rotate the axes through an angle θ:

x = X cosθ − Y sinθ, y = X sinθ + Y cosθ

Substitute into ax² + 2hxy + by² = 0

  • x² = X² cos²θ − 2XY sinθ cosθ + Y² sin²θ
  • y² = X² sin²θ + 2XY sinθ cosθ + Y² cos²θ
  • xy = X² sinθ cosθ + XY (cos²θ − sin²θ) − Y² sinθ cosθ

Collecting the coefficient of XY:

−2a sinθ cosθ + 2h (cos²θ − sin²θ) + 2b sinθ cosθ = 2h cos2θ − (a − b) sin2θ

Condition for no XY term

2h cos2θ − (a − b) sin2θ = 0

tan 2θ = 2h/(a − b), so θ = ½ tan⁻¹[2h/(a − b)]

If a = b, then cos2θ = 0 and θ = 45°.

Resulting equation

The new coefficients of X² and Y² are

λ₁, λ₂ = ½[(a + b) ± √((a − b)² + 4h²)]

and the equation becomes λ₁X² + λ₂Y² = 0 (the two axes are the bisectors of the angles between the pair of lines).

✓ Answer checked by Claude Opus 5.5 (high effort)

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