Asymptotes of Hyperbola
On this page
Direct answer
Asymptotes are the straight lines a hyperbola hugs at infinity: for x^2/a^2 − y^2/b^2 = 1 they are y = ±(b/a)x, both through the centre, with the angle between them equal to 2 tan^(-1)(b/a). When a = b the asymptotes are perpendicular, the hyperbola is rectangular (equilateral) and its eccentricity is exactly √2. For a general second-degree hyperbola, the asymptotes are found by translating to the centre (solving the two partial-derivative equations) and then setting the constant to zero — because a hyperbola and its asymptotes differ only in that constant. The conjugate hyperbola x^2/a^2 − y^2/b^2 = −1 shares the very same pair of asymptotes.
What you must remember
- Standard pair: y = ±(b/a)x through the centre; slopes ±b/a, product of slopes −b^2/a^2.
- Rectangular case: a = b gives asymptotes y = ±x, perpendicular, and e = √2; xy = c^2 has the coordinate axes themselves as asymptotes.
- Curve minus constant: after centring, the asymptotes are the same left-hand side set equal to zero — the hyperbola and its asymptotes differ only by a constant.
- Conjugate hyperbola: x^2/a^2 − y^2/b^2 = −1 shares both asymptotes; its transverse and conjugate axes swap roles.
- Angle between asymptotes: 2 tan^(-1)(b/a) for the standard curve; perpendicularity is equivalent to rectangularity is equivalent to e = √2.
- Secant behaviour: a line parallel to an asymptote meets the hyperbola in exactly one point — the second intersection has escaped to infinity.
- Method order: find the centre first, shift, then drop the constant; skipping the centre works only for origin-centred curves.
Centreing a hyperbola to find its asymptotes
Find the asymptotes of 9x^2 − 4y^2 − 18x − 16y − 43 = 0. Group the variables: 9(x^2 − 2x) − 4(y^2 + 4y) = 43. Complete both squares: 9(x − 1)^2 − 9 − 4(y + 2)^2 + 16 = 43, which tidies to 9(x − 1)^2 − 4(y + 2)^2 = 36, or (x − 1)^2/4 − (y + 2)^2/9 = 1. The centre is (1, −2), the hyperbola east-west with a = 2, b = 3. Asymptotes come from setting the left side to zero: 3(x − 1) = ±2(y + 2), giving 3x − 2y − 7 = 0 and 3x + 2y + 1 = 0. A useful cross-check: the coefficients a = 9 and b = −4 of the quadratic part deliver b^2/a^2 = 4/9 through the discriminant relation h^2 − ab = 0 − (9)(−4) > 0 confirming a hyperbola — invariants catch slips made during the completion of squares. Since a ≠ b here, the asymptotes are not perpendicular and e = √(1 + 9/4) = √13/2.
Where marks leak
JEE Main asks for asymptotes of standard or lightly shifted hyperbolas — the y = ±(b/a)x pair, read in seconds once the equation is normalised. JEE Advanced shifts the centre, rotates the frame, or asks which curves share asymptotes — the conjugate hyperbola, and the family S + λ = 0 for varying constants λ, all riding the same two lines. The recurring losses: dropping the constant term before translating to the centre, which is valid only when the centre is the origin; reporting the asymptote slope as a/b instead of b/a; and quoting e = √2 for a hyperbola that is merely translated — rectangularity depends on a = b, never on position. The secant fact (one intersection for asymptote-parallel lines) also resolves several "number of intersection points" MCQs without any computation.
Frequently asked questions
What are the asymptotes of x^2/a^2 − y^2/b^2 = 1?
The pair y = ±(b/a)x, both passing through the centre of the hyperbola.
What is the eccentricity of a rectangular hyperbola?
Exactly √2, equivalent to a = b and to perpendicular asymptotes.
Which hyperbola shares its asymptotes with a given one?
The conjugate hyperbola x^2/a^2 − y^2/b^2 = −1; the two differ only in which axis is transverse.
What are the asymptotes of xy = c^2?
The coordinate axes x = 0 and y = 0 — a rectangular hyperbola referred to its asymptotes.
How do you find asymptotes from a general second-degree equation?
Locate the centre from the partial-derivative equations, shift the origin there, then set the constant term of the centred equation to zero.