Find the sections of dfracx^24-y^2+dfracz^24=0 with the…

Find the sections of $\dfrac{x^2}4-y^2+\dfrac{z^2}4=0$ with the coordinate planes. What surface is it? Sketch it.

Explanation of Method

Recognizing a circular cone with axis Oy from its coordinate sections.

Step 1: Sections

plane	section	type
$z=0$	$y=\pm\tfrac x2$	two lines through $O$
$x=0$	$z=\pm2y$	two lines through $O$
$y=0$	$x=z=0$	the point $O$
$y=c\neq0$	$x^2+z^2=4c^2$	circle of radius $2\|c\|$

Step 2: The surface

The equation is homogeneous, and $x^2+z^2=4y^2$ means $\sqrt{x^2+z^2}=2|y|$: the distance from $Oy$ is proportional to $|y|$.

\[\text{circular cone, vertex } O,\ \text{axis } Oy,\ \tan\alpha=2\ (\alpha\approx63.4^\circ)\]

$y$ $O$

$z=0:\ y=\pm\tfrac x2$
$x=0:\ z=\pm2y$
$y=c:\ x^2+z^2=4c^2$

The cone $x^2+z^2=4y^2$ with axis $Oy$. The planes $z=0$ and $x=0$ cut it in pairs of lines; the plane $y=0$ only in the vertex $O$.

Result

\[\boxed{\,\text{circular cone, vertex } O,\ \text{axis } Oy,\ \tan\alpha=2\ (\alpha\approx63.4^\circ)\,}\]

Remark

Note: the source calls the planes $y=c$ "horizontal". They are perpendicular to the axis $Oy$, so they are vertical when $Oz$ points up. That is why the circles appear in the $xOz$ direction.