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 February 12th, 2012, 06:40 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Quadric Surfaces Reduce the equation to one of the standard forms. $x^2 - y^2 + z^2 - 2x + 2y + 8z + 12= 0$ Classify the surface. hyperbolic cylinder parabolic cylinder ellipsoid hyperbolic paraboloid cone elliptic cylinder elliptic paraboloid hyperboloid
 February 12th, 2012, 06:44 PM #2 Newbie   Joined: Aug 2010 Posts: 22 Thanks: 0 Re: Quadric Surfaces Aaron, have you tried completing the squares for x, y and z respectively? The answer should follow.
 February 12th, 2012, 06:53 PM #3 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Quadric Surfaces I get stuck when I get to the $-y^2$ term because normally when you complete the square if the leading coefficient is not one, you divide through by that number to make it one. Here I can not divide through by (-1) because it will make the other two squared terms have an coefficient of (-1) I rewrite it as: $x^2-2x-y^2+2y+z^2+8z=-12$ $x^2-2x+1 \hspace{8mm} -y^2+2y \pm ? \hspace{8mm} +z^2 +8z+16=-12+1+16+?$ $(x-1)^2 \hspace{8mm} + (y \pm ?)^2\hspace{8mm} +(z+4)^2= -12+1+16+?$
 February 12th, 2012, 07:15 PM #4 Newbie   Joined: Aug 2010 Posts: 22 Thanks: 0 Re: Quadric Surfaces Let's try: $(x^2 -2x +1) - (y^2 -2y +1) + (z^2 +8z +16)= - 12 +1 -1 +16 = 6 \Rightarrow \frac{(x-1)^2}{6} - \frac{(y-1)^2}{6} + \frac{(z+4)^2}{6} = 1$ Defining $x'= x-1, \, y' = y-1 \text{ and } z#39; = z+4$ we have $\frac{x'^2}{6} - \frac{y'^2}{6} + \frac{z'^2}{6} = 1$. We have a hyperboloid of one sheet.
 February 12th, 2012, 07:15 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Quadric Surfaces We are given: $x^2-y^2+z^2-2x+2y+8z+12=0$ Completing the square as advised: $$$x^2-2x+1$$-$$y^2-2y+1$$+$$z^2+8z+16$$=-12+1-1+16$ $(x-1)^2-(y-1)^2+(z+4)^2=2^2$ We then have a one-sheeted hyperboloid.
 February 12th, 2012, 07:19 PM #6 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Quadric Surfaces Cool, thx guys, I will do it like that next time.
 February 12th, 2012, 07:19 PM #7 Newbie   Joined: Aug 2010 Posts: 22 Thanks: 0 Re: Quadric Surfaces Oops, got a sign wrong there. Thanks Mark, fixed.

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