Reform equation for hyperbola in terms of y = f(x)

I am interested in getting the equation of a hyperbola solved for y (for up/down opening hyperbolas).

Here is how far I can get for certain

General equation: $\dfrac{(y-k)^2}{b^2}-\dfrac{(x-h)^2}{a^2}=1$

Setting h and k to zero for simplicity, $\dfrac{(y)^2}{b^2}-\dfrac{(x)^2}{a^2}=1$

then, as far as I can get with confidence in refactoring this to isolate y is

$y^2 = (a^2)(1 + \dfrac{x^2}{b^2})$

The next step is where I think I am making an error. I want the square root of $y^2$, and I get:

$y = \pm \sqrt{(a^2)(1 + \dfrac{x^2}{b^2})}$

but I think this is wrong. Can someone show me how to do this correctly?

2 Answers

$$ \frac{y^2}{b^2}-\frac{x^2}{a^2}=1\implies \frac{y^2}{b^2}=1+\frac{x^2}{a^2}\implies y^2=b^2\left(1+\frac{x^2}{a^2}\right) $$

$$\begin{align} \large\frac{y^2}{b^2}-\frac{x^2}{a^2}&=1\\ \large y^2a^2-x^2b^2&=a^2b^2\\ \large y^2a^2&=x^2b^2+a^2b^2\\ \large y^2&=\frac{x^2b^2}{a^2}+b^2\\ \large y&=\pm\sqrt{\frac{x^2b^2}{a^2}+b^2}\\ \large y&=\pm\sqrt{b^2(\frac{x^2}{a^2} +1)}\\ \end{align}$$