Name the curve

Assuming that $c \ne 0$ is a constant, we can rewrite the given equation as

$\dfrac{y + x}{xy} = \dfrac{1}{c} \Longrightarrow c(y + x) = xy \Longrightarrow xy - cx - cy = 0 \Longrightarrow (x - c)(y - c) = c^{2}$ .

Letting $x' = x - c$ and $y' = y - c$ the equation becomes $x'y' = c^{2}$ , which is the equation of a rectangular $\boxed{\text{hyperbola}}$ with axes rotated $45^{\circ}$ .

To see how this can be converted to the standard form of a hyperbola, let $X = \dfrac{x' + y'}{\sqrt{2}}$ and $Y = \dfrac{-x' + y'}{\sqrt{2}}$ . Then

$x' = \dfrac{X - Y}{\sqrt{2}}$ and $y' = \dfrac{X + Y}{\sqrt{2}} \Longrightarrow c^{2} = x'y' = \dfrac{X^{2}}{2} - \dfrac{Y^{2}}{2}$ ,

i.e., a rectangular hyperbola in the $XY$ -coordinate system where $X = (x - c)\cos(45^{\circ}) + (y - c)\sin(45^{\circ})$ and $Y = -(x - c)\sin(45^{\circ}) + (y - c)\cos(45^{\circ})$ , which represents a $45^{\circ}$ rotation from the standard $xy$ -axes and a shift of $\sqrt{2}c$ upward along the line $y = x$ .