Answer
The surface is an elliptic paraboloid with the x-axis as its axis and the origin as its vertex.
Work Step by Step
Setting $y$=0,1,2,...k,... the traces are parabolas$, x=z^{2}+k^{2}.$
Setting $z$=0,1,2,...,k,... the traces are parabolas$, x=y^{2}+4k^{2}$
Setting x=0, we have
$ y^{2}+4z^{2}=0\qquad$ , producing a single point (the origin).
Setting x$ < 0$, we have
$ y^{2}+4z^{2} < 0\qquad$ , no such points.
The curve has no traces in planes $x=k$ where k is negative.
Setting x=1,
$y^{2}+4z^{2}=1$
$\displaystyle \frac{y^{2}}{1}+\frac{z^{2}}{(1/2)^{2}}=1\qquad$ , producing an ellipse, a=1, b=1/2.
Setting x=$4$,
$y^{2}+4z^{2}=4$
$\displaystyle \frac{y^{2}}{2^{2}}+\frac{z^{2}}{(1)^{2}}=1\qquad$ , producing an ellipse, a=2, b=1.
The rectangle about the ellipse increases in size.
Setting x=$9$,
$y^{2}+4z^{2}=9$
$\displaystyle \frac{y^{2}}{3^{2}}+\frac{z^{2}}{(3/2)^{2}}=1\qquad$ , producing an ellipse, a=$3$, b=$3/2$.
The rectangle about the ellipse keeps increasing in size.
The surface is an elliptic paraboloid with the x-axis as its axis and the origin as its vertex.
