Complete square of hyperbola to find asymptotes?
- Thread starter svextra
- Start date
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).