1. Reducing Quadratic Surface Equation, HELP!

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!

2. 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!
$(x^2-2x) - (y^2 -2y) + (z^2+8z) + 13 = 0$
$((x-1)^2 - 1) - ((y-1)^2 - 1) + ((z+4)^2 - 16) + 13 = 0$

3. Hello,
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

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standard form of surface equation

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