Question

Check whether the surface represented by
x^2 + y^2 = 2 (z + 1) is symmetric about the YZ-plane, ZX-plane and XY-plane. Do the coordinate axes intersect the surface?

Answer 1

Solution. Considering this equation, we see that the surface x^2 + y^2 = 2 (z + 1) is an elliptic paraboloid. Find the intersection of the surface of the planes parallel to the XY-plane (z=h, where h is constant). Get

$$x^2+y^2=2(h+1)$$

For h=-1 get point (0,0,-1); for h>-1 get circle x^2+y^2=2(h+1). For XY-plane (h=0) get x^2+y^2=2. Therefore the surface intersects the coordinate axes 0x and 0y at the points:

$$0x: (-\sqrt2, 0,0) and (\sqrt2, 0,0)$$

$$0y: (0,-\sqrt2,0) and (0, \sqrt2,0)$$

The surface relative to the XY-plane is not symmetrical.

Find the intersection of the surface of the planes parallel to the YZ-plane (x=h, where h is constant). Get

$$y^2=2z+2-h^2.$$

This equation is a parabola equation. For YZ-plane (x=0) get y^2 = 2 (z + 1). The top of the parabola at the point (0,0,-1). Parabola intersects the coordinate axis 0y at points:

$$0y: (0,-\sqrt2,0) and (0, \sqrt2,0)$$

Considering the symmetric planes x = -h and x = h, we get the same parabolas y^2 = 2 (z + 1)-h^2. Hence the surface relative to the YZ-plane is symmetrical.

Find the intersection of the surface of the planes parallel to the ZX-plane (y=h, where h is constant). Get

$$x^2=2z+2-h^2.$$

This equation is a parabola equation. For ZX-plane (y=0) get x^2 = 2 (z + 1). The top of the parabola at the point (0,0,-1). Parabola intersects the coordinate axis 0y at points:

$$0x: (-\sqrt2, 0,0) and (\sqrt2, 0,0)$$

Considering the symmetric planes y = -h and y = h, we get the same parabolas x^2 = 2 (z + 1)-h^2. Hence the surface relative to the ZX-plane is symmetrical.

Answer. The surface is symmetric about YZ-plane and ZX-plane. The surface relative to the XY-plane is not symmetrical.

The surface intersects the coordinate axes in points:

$$0x: (-\sqrt2, 0,0) and (\sqrt2, 0,0)$$

$$0y: (0,-\sqrt2,0) and (0, \sqrt2,0)$$

$$0z: (0,0,-1)$$