Kick in the Dick Conics

Identify the conic. Then complete the square to write the conic in standard form, state the coordinates of the center and foci, solve for y and sketch the graph.

12x^(2)-4y^(2)-72x-16y+44=0

Here's what I've done so far:

12x^(2)-4y^(2)-72x-16y+44=0

12x^(2)-72x-4y^(2)-16y=-44

12(x^(2)-6x+9)-4y^(2)-16y=-44+0+9*12

12((x-3)^(2))-4y^(2)-16y=-44+0+9*12

12(x-3)^(2)-4y^(2)-16y=-44+0+9*12

12(x-3)^(2)-4(y^(2)+4y+4)=-44+0+9*12+0+4*-4

12(x-3)^(2)-4((y+2)^(2))=-44+0+9*12+0+4*-4

12(x-3)^(2)-4(y+2)^(2)=-44+0+9*12+0+4*-4

12(x-3)^(2)-4(y+2)^(2)=-44+0+108+0-16

12(x-3)^(2)-4(y+2)^(2)=48

((x-3)^(2))/(4)-((y+2)^(2))/(12)=1

What's going on here? My instructor told us we'd be able to just figure it out but I don't see shit about it in the textbook that's worth anything.

This one asks us to solve for y and graph it:

10x^(2)-8xy+6y^(2)+8x-5y-30=0

-8xy+6y^(2)-5y=-10x^(2)-8x+30

-8xy+6y^(2)-5y+10x^(2)+8x-30=0

6y^(2)-8yx-5y=-10x^(2)-8x+30

y=-10x^(2)-8x+30

Seem right?

I dunno

Google quadratic surfaces.  Your search to enlightenment begins there.

It's a hyperbola. Center is (3, -2). I don't remember what the foci is, but I'm sure google can teach you that.

>>4
Where do you get hyperbola? I see no asymptotes. Looks more like an ellipse.

ax^2 - by^2 = c is going to give you a hyperbola

ax^2 + by^2 = c is going to give you an ellipse

>>6
Right, my bad.