Conic Sections: Graphing with Multiple Squares

[duki]

Homework Statement

graph the following

Homework Equations

$$9x^2+4y^2+36x-8y+4=0$$

The Attempt at a Solution

I think I need to get it into $$\frac{(x-x0)^2}{a^2}+\frac{(y-y0)^2}{b^2}$$ but I'm not sure.
I have $$\frac{9x^2}{-4}-8x+y^2-2y=1$$ and now I'm stuck

 

[duki]

Ok, update. Here's what I have so far:

$$\frac{(x+2)^2}{2^2}+\frac{(y-1)^2}{3^2} = 1$$

Does that look right?

 

[Dick]

It looks fine, if you meant 4y^2 in the original post and not 47^2.

 

[duki]

Thanks for catching that. I fixed it.
Ok, so now I have the following:

$$Center = (-2,1)$$
$$a = 2$$
$$b = 3$$
$$Verticies: (0,1),(-4,1),(-2,4),(-2,-2)$$

Does that look right?

 

[Dick]

Looks right to me. But I'm tired. You should double check.

 

[duki]

haha. you're supposed to be the double checker!

I'm really stuck now. I'm trying to find 'c' and I get $$\sqrt{-5}$$. Did I do something wrong? c is the square root of a^2 - b^2 right? Here, a = 2 and b = 3. I'm confused

 

[Dick]

No. YOU are supposed to be the double checker. It's your class. I'm just tossing off hints without being fully awake. I have no idea what 'c' is supposed to be. Could you just like say what it is supposed to be instead of dropping a cryptic letter? I'll take another guess and say 'distance from center to focus'? That's a lot better description than 'c'. Why don't you think it could be sqrt(3^2-2^2)? If you flip the x and y axes, do you think this distance should change from real to imaginary?

 

[duki]

hmm, I'm not sure. I didn't know you could swap them like that. The formula I was going by said "distance from center to focus" = sqrt(a^2 - b^2). If you flip them, you get the real answer?

 

[HallsofIvy]

Surely that formula was assuming a> b. That is, that a is the length of the longer semi-axis. Don't just memorize formulas. Learn what the mean.