Complete square of hyperbola to find asymptotes?

[svextra]

Equation given is: 2y^2 + 4y - 4x^2 = 0
I don't think i'm doing it right because I end up with 2(x)^2
I also do not know how to find the asymptotes from this equation.

My work is below.

 

[Ishuda]

Equation given is: 2y^2 + 4y - 4x^2 = 0
I don't think i'm doing it right because I end up with 2(x)^2
I also do not know how to find the asymptotes from this equation.

My work is below.

The equation for a hyperbola is
$\displaystyle \left[\dfrac{u}{a}\right]^2\, -\, \left[\dfrac{v}{b}\right]^2\, =\, 1$
which is what you have but in a slightly different form [u = y+1, a=1, v = x, b = $\displaystyle \frac{1}{\sqrt{2}}$]. To find the asymptotes, we first rearrange the equation to
$\displaystyle \left[\dfrac{v}{u}\right]^2\, =\, \left[\dfrac{b}{a}\right]^2\, - \left[\dfrac{b}{u}\right]^2$

note that when u² becomes 'very large', v² becomes 'very large and (b/u)² become very small. So
$\displaystyle v\,$~$\displaystyle \, \pm\dfrac{b}{a}\, u$
and we have the asymptotes as $\displaystyle \pm$(b/a).