-x^2-4x+4y^2+8y-22=0

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- Nov 12th 2008, 06:59 PMscottydointIs this problem a general for of the equation of a circle?
-x^2-4x+4y^2+8y-22=0

- Nov 12th 2008, 07:27 PMSoroban
Hello, scottydoint!

Quote:

$\displaystyle -x^2-4x+4y^2+8y-22\:=\:0$

It is not a circle . . . $\displaystyle x^2$ and $\displaystyle y^2$ have different coefficient.

Since the coefficients of $\displaystyle x^2$ and $\displaystyle y^2$ have__opposite__signs, it is a**hyperbola**.

- Nov 12th 2008, 07:38 PMearboth
1. Complete the squares:

$\displaystyle -(x^2+4x{\bold{\color{red}+4}})+4(y^2+2y{\bold{\col or{blue}+1}})=22 {\bold{\color{red}-4}} {\bold{\color{blue}+4}}$

2. Re-write:

$\displaystyle -x^2-4x+4y^2+8y-22=0~\implies~ -\dfrac{(x+2)^2}{22} + \dfrac{(y+1)^2}{5.5} =1$

3. This is the equation of a hyperbola opening up-down.