Asymptotes of (x²−y²)²−2(x²+y²)+x−1=0

Short answer

The asymptotes are y = x + 1, y = x - 1, y = -x + 1, y = -x - 1. Double roots m=±1 from φ₄(m)=0 lead to equation 4c²-4=0, giving c=±1 for each slope.

Find the asymptotes of the curve (x²−y²)² − 2(x²+y²) + x − 1 = 0.

Curve: (x² − y²)² − 2(x² + y²) + x − 1 = 0. Degree 4.

φ₄(m) = (1 − m²)²
φ₃(m) = 0
φ₂(m) = −2(1 + m²)
φ₁(m) = 1

φ₄(m) = 0 gives m = 1 (twice) and m = −1 (twice). Since φ₃ = 0, for a double root use

c² φ₄″(m)/2 + c φ₃′(m) + φ₂(m) = 0.

φ₄ = (1 − m)²(1 + m)², so at m = 1: φ₄″/2 = (1 + m)² = 4. At m = −1: φ₄″/2 = (1 − m)² = 4.

m = 1: 4c² + 0 − 4 = 0, so c = ±1. Asymptotes: y = x + 1, y = x − 1.

m = −1: 4c² + 0 − 4 = 0, so c = ±1. Asymptotes: y = −x + 1, y = −x − 1.

Answer: y = x ± 1 and y = −x ± 1 (four asymptotes).

✓ Answer checked by Claude Opus 5.5 (high effort)

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