# Reducing Quadratic Surface Equation, HELP!

• Feb 9th 2009, 01:57 PM
acg716
Reduce the equation to one of the standard forms

x2 – y2 + z2 – 2x + 2y + 8z + 13 = 0

And what kind of surface is it?

I got (x-1)^2 - (y-1)^2 + (z+4)^2 = 5
that answer was wrong, so then i switched the (y-1)^2 to (y+1)^2 and it was still wrong. I also used complete the square to solve it. So now I'm lost and not sure what to do, HELP!
• Feb 9th 2009, 08:37 PM
ThePerfectHacker
Quote:

Originally Posted by acg716
Reduce the equation to one of the standard forms

x^2 – y^2 + z^2 – 2x + 2y + 8z + 13 = 0

I got (x-1)^2 - (y-1)^2 + (z+4)^2 = 5
that answer was wrong, so then i switched the (y-1)^2 to (y+1)^2 and it was still wrong. I also used complete the square to solve it. So now I'm lost and not sure what to do, HELP!

$(x^2-2x) - (y^2 -2y) + (z^2+8z) + 13 = 0$
$((x-1)^2 - 1) - ((y-1)^2 - 1) + ((z+4)^2 - 16) + 13 = 0$
• Feb 11th 2009, 10:33 AM
Moo
Hello,
Quote:

Originally Posted by acg716
Reduce the equation to one of the standard forms

x^2 – y^2 + z^2 – 2x + 2y + 8z + 13 = 0

I got (x-1)^2 - (y-1)^2 + (z+4)^2 = 5
that answer was wrong, so then i switched the (y-1)^2 to (y+1)^2 and it was still wrong. I also used complete the square to solve it. So now I'm lost and not sure what to do, HELP!

I get (x-1)²-(y-1)²+(z+4)²=3

We have (by ordering) :
$(x^2-2x)-(y^2-2y)+(z^2+8z)+13=0$
$[(x-1)^2-1]-[(y-1)^2-1]+[(z+4)^2-16]+13=0$
$(x-1)^2-1-(y-1)^2{\color{red}+}1+(z+4)^2-16+13=0$

which gives my solution (Tongueout)