Why is the equation $z=(x+y)^2+y^2$ a paraboloid?

Question

Why is the equation $z=(x+y)^2+y^2$ a paraboloid?, as the right side is expanded, it is $z=x^2+2xy+2y^2$.

The equation of a paraboloid in some sources is simply $z=\frac{x^2}{a^2}+\frac{y^2}{b^2}$ and there is no the product of $x$ and $y$ in the equation.

Answer 1

A paraboloid is something which can be described by such an equation after possibly making a change of coordinate. On the right hand side, if you replace $x = x' - y$ you get the standard form, $z= (x')^2 + y^2$.

Answer 2

The equation $z={x^2\over a^2}+{y^2\over b^2}$ isn’t the equation of a paraboloid, any more than ${x^2\over a^2}+{y^2\over b^2}=1$ is the equation of an ellipse in the $x$-$y$ plane. Those sources you quote likely mentioned somewhere that this is the equation of a paraboloid in “standard position,” that it’s the “standard form” of the equation, or some other wording along those lines. It describes an elliptic paraboloid that has its vertex at the origin, the coordinate axes as its principal axes, and opens “upward” in the direction of the positive $z$-axis. Often, we also assume that $a\ge b$. This is a very special subset of all possible elliptic paraboloids.

The paraboloid that you have has been rotated approximately 32° from this standard position, which you can see if you plot the surface. If you were to apply the inverse rotation to it, you’d end up with an equation in the standard form. We don’t need to do this, however, as the original form of the equation of this paraboloid makes it easy to find its spectrum, which can be used to distinguish among the various types of quadric surfaces: it’s $(1,1,0,-1)$, which is the spectrum that all elliptic paraboloids have. (Compare this to, say, the spectrum of an ellipsoid, which from the standard equation is $(1,1,1,-1)$.)

Answer 3

You may get rid of the term in $xy$ by performing the transformation $$x=u+\frac{1+\sqrt 5}{2}v,\,y=\frac{1+\sqrt 5}{ 2}u-v,$$ which gives you an equation in standard form.