-x^2-4x+4y^2+8y-22=0
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-x^2-4x+4y^2+8y-22=0
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.
1. Complete the squares:Quote:
Originally Posted by scottydoint
$\displaystyle -(x^2+4x{\bold{\color{red}+4}})+4(y^2+2y{\bold{\col or{blue}+1}})=22 {\bold{\color{red}-4}} {\bold{\color{blue}+4}}$
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.
