Rectangular Hyperbola
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Direct answer
xy = c² is the rectangular (equilateral) hyperbola in its most exam-friendly dress: the coordinate axes themselves are the asymptotes, perpendicular to each other — the defining mark, equivalent to eccentricity exactly √2. The parametric point is (ct, c/t); the foci sit on y = x at (c√2, c√2) and its negative — rotate the standard form x² − y² = 2c² to see it, with transverse axis 2c√2 and latus rectum equal to it because b = a. The standard-form version goes rectangular precisely when b = a, asymptotes y = ±x. Its cleanest identity is arithmetic: the product of the coordinates is constant, so (2, 4) and (8, 1) both ride the curve xy = 8.
What you must remember
- Canonical form: xy = c² with asymptotes the coordinate axes; parametric point (ct, c/t), and the parameter is the reciprocal partner of the coordinate ratio.
- Eccentricity: e = √2 always — the only conic whose eccentricity is fixed with no free parameter deciding it.
- Foci and vertices: on y = x for xy = c² — vertices (±c, ±c) and foci (±c√2, ±c√2); for xy = 8, that reads vertices (±2√2, ±2√2) and foci (±4, ±4).
- Standard-form twin: x² − y² = a² is rectangular because b = a; asymptotes y = ±x; same curve as xy = a²/2 rotated by 45°.
- Latus rectum equals transverse axis: 2b²/a = 2a here, a fact unique to the rectangular case.
- Tangent at parameter t: x + t²y = 2ct; at a point (x1, y1) on xy = c², the tangent is xy1 + yx1 = 2c².
- Director circle degenerates: x² + y² = a² − b² = 0, a single point — no real perpendicular tangent pairs exist off the centre.
- Conjugate curiosity: the conjugate of a rectangular hyperbola is congruent to it, just rotated into the other pair of quadrants.
Rotating xy = c²
Work xy = 8 completely. Rotate axes by 45°: X = (x + y)/√2, Y = (y − x)/√2 turns xy = (X² − Y²)/2, so the equation becomes X² − Y² = 16 — a standard hyperbola with a² = 16, b² = 16, hence e = √(1 + 1) = √2. Its vertices are (±4, 0) in (X, Y), which rotate back to (4/√2, 4/√2) = (2√2, 2√2) and its negative; its foci sit at (±ae, 0) = (±4√2, 0), returning as (4, 4) and (−4, −4). The transverse axis lies along y = x with length 8, and the latus rectum is also 8 — the rectangular privilege. Check the tangent machinery at the point (2, 4): the tangent reads x(4) + y(2) = 16, or 2x + y = 8, and indeed the point satisfies both the curve (2 × 4 = 8) and the tangent. The rotation is the entire theory of this curve: every formula from the standard hyperbola is available, then translated back through 45°.
Foci that sit on y = x
JEE Main asks the identifier facts: eccentricity √2, the asymptotes of xy = c², and the foci positions — with the standing distractor placing the foci on the coordinate axes as if the curve were standard-form. A second Main pattern gives xy = 16 or x² − y² = 8 and asks for vertices or axis lengths — one-liners once the rotation is internalised. JEE Advanced prefers the constructive side: tangents and normals at parameter t (x + t²y = 2ct and its normal counterpart), chords with a given midpoint through the T = S₁-style substitution, and the locus family — the general second-degree equation ax² + 2hxy + by² + ... represents a rectangular hyperbola exactly when a + b = 0 and h² > ab, a coefficient test worth memorising. The recurring candidate error is algebraic rather than conceptual: in xy = c², the parameter satisfies x = ct, y = c/t, and mixing the two produces points that never lived on the curve. Verify any claimed parametric point by multiplying the coordinates back to c².
Frequently asked questions
What makes a hyperbola rectangular?
Perpendicular asymptotes — in standard form x²/a² − y²/b² = 1 this means b = a, and the eccentricity is then exactly √2.
Where are the foci of xy = c²?
On the line y = x, at (c√2, c√2) and (−c√2, −c√2) — for xy = 8, that is (4, 4) and (−4, −4).
What is the parametric point on xy = c²?
(ct, c/t): the coordinates multiply to c² for every parameter value t.
What is the tangent at the point (x1, y1) on xy = c²?
xy1 + yx1 = 2c² — the same substitution pattern the parabola and ellipse use for tangents at a point.
How do you test whether a general second-degree equation is a rectangular hyperbola?
Check a + b = 0 together with h² > ab (with the usual conic non-degeneracy) — the coefficient signature of perpendicular asymptotes.