So, you know that the equation is going to look like
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$$
You already figured out that $a=-b$. So now you need to figure out what values of $a$ and $b$ will make the foci be at the given point $(4,0)$ and $(-4,0)$.
Since you know that $a=-b$, then you know that $a^2=b^2$, so you can rewrite the equation as
$$\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$$
or, by clearing denominators (multiplying through by $a^2$)
$$x^2 - y^2 = a^2.$$
Now: where does the hyperbola intersect the $x$ axis? If you have a point $(X,0)$ in the hyperbola, then plugging it in you will have
$$X^2 = a^2.$$
So if you can determine where the hyperbola intersects the $x$-axis, you can use that to figure out the value of $a$, and so the value of $b$, and so the equation for the hyperbola.
Note: There are many, many, many ways to finish the problem, depending on what you know about hyperbolas in general (which is precisely why I asked you in comments what you knew).
Added. One possible way to finish the problem is to use the eccentricity of the hyperbola. The distance from the center to the foci is $a\varepsilon$, where $\varepsilon$ is the eccentricity of the hyperbola. The eccentricity is always equal to
$$\varepsilon = \sqrt{ 1 + \frac{b^2}{a^2}}$$
so here, since $a=b$, that means that the eccentricity is $\sqrt{2}$. Therefore, since the distance from the center to the foci is $4$, this tells you that $4 = a\varepsilon = a\sqrt{2}$. You now know what $a$ is, and thus, you know what $a^2=b^2$ is, and so you know the equation of the hyperbola.
Added 2. Another possibility you discuss in the comments. You have $a$ is the distance from the center to the vertices; $b$ is the distance from the center to the conjugate axis. These distances satisfy $c^2 = a^2+b^2$, where $c$ is the value $4$ that you have. Since $a^2=b^2$, then you have $c^2=2a^2$, or $c = \sqrt{2a^2} = \sqrt{2}|a| = a\sqrt{2}$. You already know what $c$ is, so now you can solve for $a$. Once you know $a$, you also know $b$ (since $a=-b$) which gives you the information you need.
Just to elaborate on Michael Hardy's answer a little bit...
Suppose we have a rational function written like $f(x)=\frac{p(x)}{q(x)}$. Performing the long division $p(x)\div q(x)$ gives you a different way of writing the same function, $$f(x)=a(x)+\frac{r(x)}{q(x)},$$ as a polynomial plus a rational function whose numerator's degree is smaller than its denominator's degree. (To be more specific, $a(x)$ has degree equal to the degree of the original numerator $p(x)$ minus the degree of the denominator $q(x)$. The degree of $r(x)$ is smaller than that of $q(x)$ because $r(x)$ is the remainder of the long division.)
As you already understand, $\frac{r(x)}{q(x)}$ tends to zero as $x$ tends to infinity, because the degree of the numerator is smaller than the degree of its denominator. That means that as $x\rightarrow\infty$, the function $f(x)$ tends towards the polynomial $a(x)$.
If $p(x)$ has degree exactly one more than that of $q(x)$, then $a(x)$ has degree one: it's a line. We call this line a slant or oblique asymptote of $f(x)$. If the degree of $p(x)$ is more than one higher than that of $q(x)$, then $a(x)$ is a quadratic, or a cubic, or whatever, but it's not a line. So $f(x)$ doesn't have a slant asymptote in this direction. (We still might say something like $f(x)$ asymptotically approaches a(x).)
Hopefully this explains why asymptotes only occur when the degree of the numerator is exactly one more than that of the denominator. It also might give you a hint for how you can find slant asymptotes of functions that aren't rational: if you can rewrite your function as a line plus something that goes to zero, you've got yourself an asymptote!
Best Answer
Edited to do it properly -- see below
Original post:
We have $$y=b\sqrt{\frac{x^2}{a^2}-1}=\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}$$ And as $x\to\pm\infty$, $\sqrt{1-\frac{a^2}{x^2}}\to 1$.
End of original post
But as mentioned in the comments, it is not enough to show that $\frac{y}{bx/a}\to 1$. We have to show that $y-\frac{b}{a} x\to 0$:
$$y-\frac{b}{a}x=\frac{b}{a}x\left(\sqrt{1-\frac{a^2}{x^2}}-1\right)$$ But $$1-\frac{a^2}{x^2}\le\sqrt{1-\frac{a^2}{x^2}}<1$$ So $$\left|\sqrt{1-\frac{a^2}{x^2}}-1\right|\le\frac{a^2}{x^2}$$ Therefore $$\left|y-\frac{b}{a}x\right|\le\frac{b}{a}|x|\cdot\frac{a^2}{x^2}=\frac{ba}{|x|}$$ which tends to $0$ as $x\to\pm\infty$.