Question

i got this equation x^2-y^2+z^2-4x-2y-2z+4=0, i reduced it to (x-2)^2-(y+1)^2+(z-1)^2=0. Wolfram told me this is an infinite cone, how do i go about graphing it?

Answer 1

No idea, but I'd like to bookmark the question.

Answer 2

you know what, i think i know how to do it. This is an infinite cone, but its shifted, meaning its not on the origin.

Answer 3

so, i suppose i could set z=0

Answer 4

then obtain the lines, that would form the cone itself

Answer 5

What does this cone look like:
x^2 + z^2 - y^2 = 0 ?

It's a cone that expands along the y-axis and has its apex at the origin.

Answer 6

right, no problem

Answer 7

Similarly, your cone expands parallel to the y-axis and has its apex at (2,-1, 1)

Answer 8

in this case the cone would open on the y-axis cause of the negative sign

Answer 9

very good

Answer 10

Yes

Answer 11

Notice also by the way that the cone is also 'double-sided'; it open up in both directions from the apex.

Answer 12

so, in reality, there is no need to find the traces, it would appear to be enough to draw intersecting lines through the apex

Answer 13

i get a circle , with x=0;yz-plane
also another circle with y=0;xz-plane
then with z=0, there is no graph

Answer 14

All the cross sections in the xz-plane--for y = constant--will be circles. Stick for the moment with the case
x^2 + z^2 - y^2 = 0

For EVERY value of y except y = 0, the locus of the xz points in the xz plane will be circle because the equation above is equivalent to
\[ x^2 + z^2 = y^2 > 0 \]
Hence if y is a non-zero constant, then the locus of the points (x,y,z) which satisfy that equation is a circle in the xz plane.