How to calculate the asymptotes to $y^2+2y-4x^2=0$

Question

Calculate the center and asymptotes to the hyperbola $y^2+2y-4x^2=0$, aswell as the intersections of the actual coordinate axis.

The hyperbola takes the form $\left ( \frac{x}{\frac{1}{\sqrt{4}}}\right )^{2}-\left ( {y+1}\right )^{2}=-1$ and I understand how to complete all of the steps except how to find the asymptotes to the hyperbola. My course textbook states that they should be given by the formula $y=\pm \frac{b}{a}x$ if $(\frac {y}{b}^{2})=(\frac{x}{a})^{2}\pm1\approx (\frac{x}{a})^2$. But if I use this formula I get that $y+1=\pm \sqrt4 x $. What did I do wrong?

Answer 1

Others have done much the same, but I would do this a little differently.

$$(y+1)^2-4x^2=1$$means we have the hyperbola $$(y+1-2x)(y+1+2x)=1$$

If we remember that the asymptotes to $xy=1$ are $x=0$ and $y=0$ the condition for the asymptote is simply that one of the factors is zero, and you have your two lines.

Whether that is up to the standard of rigour you need, I don't know, but once you know the asymptotes it is easy to prove they are right if you need to show that the curve comes arbitrarily close.

Answer 2

I think your hyperbola has the form \begin{align*}y^2 + 2y {\color{red}+\color{red}1\color{red}-\color{red}1}-4x^2 =0 &\Leftrightarrow (y+1)^2 -1- 4x^2= 0 \\&\Leftrightarrow (y+1)^2 - 4x^2=1 \end{align*}

So $a = \frac{1}{2}$ and $b= 1$.

Then you should get the asymptotes $$y = \pm 2x-1$$

Answer 3

You should consider the center of hyperbola,$(0,-1)$, in fact: $$\frac{\pm x}{1 / \sqrt 4}=y+1$$ are the true asymptotes.

Answer 4

Rewrite the formula \begin{eqnarray*} (y+1)^2= \color{red}{1} +(2x)^2. \end{eqnarray*} Now for large $x$ and $y$ the $\color{red}{1}$ is negligible and we have $ y+1 = \pm 2x$.