Geometrical interpretation of $P(x) + Q(y) = 0 $ when P,Q are polynomials of degree 2? Special cases are circles ( $ (x-x_0)^2 + (y-y_0)^2 = R $ ) and ellipses. 
Is there a geometric interpretation in the general case $ ( ax^2 + bx + c ) + ( dy^2 + ey + f ) = 0$?
 A: They are called conics.  They also include parabolas $x^2+y=0$, pairs of lines $x^2-y^2=0$, hyperbolas $x^2-y^2=1$, single lines $x^2=0$, single points $x^2+y^2=0$ and the empty set $x^2+y^2=-1$.  Usually, one can have a seventh term $gxy$; when you have that, the shapes rotate, and are not vertically symmetric.
A: By completing the square, you find that
$$
(ax^2 + bx + c) + (dy^2 + ey + f) = (a(x- \alpha)^2 + s) + (d(y - \beta)^2 + t))
$$
so that your equation has the form:
$$
a (x - \alpha)^2 + d (y - \beta)^2 = \kappa.
$$
Assume first that $a, d > 0$ (or both are negative).
If $\kappa < 0$, then there are no solutions, while if $\kappa = 0$, then the only solution is the point $(\alpha,\beta)$.
The interesting case is when $\kappa > 0$.  Then you get a circle if $a = d$ and an ellipse otherwise, centered at the point $(\alpha, \beta)$.
Similar analysis applies to other cases.  If $a$ and $d$ are both non-zero, but have opposite signs, then you get a hyperbola.  If either $a$ or $d$ is zero but the other is not, you get a parabola (or two lines or a double line, in degenerate cases).  If both $a$ and $d$ are zero, then you get a line (or no solutions in a degenerate case).
