- 2071 Shrawan · 5 marks
Asymptotes of a²y²+x²y²−a²x²+2ax³−x⁴=0
Short answer
Set highest-degree terms: slopes m = ±1. Then c = a/m gives asymptotes y = x − a and y = −x + a.
Find the asymptotes of the curve a²y² + x²y² − a²x² + 2ax³ − x⁴ = 0.
Curve: a²y² + x²y² − a²x² + 2ax³ − x⁴ = 0. Degree 4.
Parallel to the axes: the coefficient of y² is a² + x², which never vanishes for real x, so no asymptote parallel to the y-axis. The highest power of x is x⁴ with a constant coefficient (−1), so no asymptote parallel to the x-axis.
Oblique asymptotes:
φ₄(m) = m² − 1
φ₃(m) = 2a
φ₂(m) = a²m² − a²
φ₄(m) = 0 gives m = 1 and m = −1 (both simple). With c = −φ₃(m)/φ₄′(m) and φ₄′ = 2m:
- m = 1: c = −2a/2 = −a. Asymptote: y = x − a.
- m = −1: c = −2a/(−2) = a. Asymptote: y = −x + a.
(Check: the curve is y² = x²(x − a)²/(x² + a²), so y = ±x(x − a)/√(x² + a²) ≈ ±(x − a) for large x.)
Answer: y = x − a and y = −(x − a), i.e. x − y = a and x + y = a.
✓ Answer checked by Claude Opus 5.5 (high effort)
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