Rectangular Hyperbola

Special case when the asymptotes are perpendicular (a = b, eccentricity e = √2). The form xy = c² arises from x²/c² − y²/c² = 1 rotated 45°. Asymptotes are the coordinate axes.

A rectangular hyperbola has perpendicular asymptotes ($a = b$). Starting from $X^2/a^2 - Y^2/a^2 = 1$ (asymptotes $Y = \pm X$), rotate axes by 45°:

$$X = \frac{x-y}{\sqrt{2}}, \quad Y = \frac{x+y}{\sqrt{2}}$$

Substituting:

$$\frac{(x-y)^2/2}{a^2} - \frac{(x+y)^2/2}{a^2} = 1$$ $$\frac{(x-y)^2 - (x+y)^2}{2a^2} = 1 \implies \frac{-4xy}{2a^2} = 1 \implies xy = -\frac{a^2}{2}$$

Setting $c^2 = a^2/2$ (and noting the sign depends on branch convention):

$$xy = c^2$$

Properties:

• Asymptotes: the coordinate axes ($x = 0$ and $y = 0$)
• Eccentricity: $e = \sqrt{2}$ (since $a = b$ gives $e = \sqrt{1+b^2/a^2} = \sqrt{2}$)
• Foci: at $(c\sqrt{2}, c\sqrt{2})$ and $(-c\sqrt{2}, -c\sqrt{2})$
• Two branches: first quadrant ($x > 0, y > 0$) and third quadrant ($x < 0, y < 0$)

Director circle: $x^2+y^2 = a^2-b^2 = 0$ — the centre is the only point from which perpendicular tangents can be drawn.

The curve is its own conjugate: The conjugate of $xy = c^2$ is $xy = -c^2$ (the other two quadrants), sharing the same asymptotes.