For simplicity, let's say that the hyperbola is centered at $\left(0\, 0\right)$ with the following foci: E at $\left(-c\, 0\right)$ and F at $\left(c\, 0\right)$. So, the distance from each focus to the center is c.

The distance from a general point $P\left(x\, y\right)$ to E is given by $\mathrm{PE} \= \sqrt{{\left(x - \left(-c\right)\right)}^2\+{\left(y - 0\right)}^2} \= \sqrt{{\left(x\+c\right)}^2\+y^2}$.

The distance from P to F is given by $\mathrm{PF} \= \sqrt{{\left(x - c\right)}^2\+{\left(y -0\right)}^2}\= \sqrt{{\left(x\mathit{ }\mathit{-}\mathit{ }c\right)}^2\+y^2}$.

Looking at the case in which P is a vertex of the hyperbola and subtracting the distances from this vertex to each focus, we see that the difference of these distances is $2a$.

So, we know that:         

$\mathrm{PE}-\mathrm{PF}\=2 a$

$\sqrt{{\left(x\+c\right)}^2\+y^2}-\sqrt{{\left(x-c\right)}^2\+y^2}\=2a$

$\sqrt{{\left(x\+c\right)}^2\+y^2} \= 2a \+ \sqrt{{\left(x-c\right)}^2\+y^2}$

${\left(x\+c\right)}^2\+y^2 \= {\left(2 a \+ \sqrt{{\left(x-c\right)}^2\+y^2}\right)}^2$

$x^2\+2cx\+c^2\+y^2\=4a^2\+4 a\sqrt{{\left(x-c\right)}^2\+y^2}\+{\left(x-c\right)}^2\+y^2$

$x^2\+2cx\+c^2\+y^2 \= 4a^2\+4 a\sqrt{{\left(x-c\right)}^2\+y^2} \+ x^2 - 2 c x \+ c^2 \+ y^2$

$2c x \= 4 a^2\+4 a\sqrt{{\left(x-c\right)}^2\+y^2} - 2 c x$

$4c x - 4 a^2 \= 4 a \sqrt{{\left(x-c\right)}^2\+y^2}$

$c x - a^2 \= a \sqrt{{\left(x-c\right)}^2\+y^2}$

${\left(c x - a^2\right)}^2 \= a^2\left({\left(x-c\right)}^2\+y^2\right)$

$c^2{x}^2-2a^{2}cx\+ a^4 \= a^2{x}^2-2 a^{2}cx\+ a^2{c}^{2 }\+a^2{y}^2$

$c^2{x}^2\+ a^4 \= a^2{x}^2\+ a^2{c}^{2 }\+a^2{y}^2$

$a^4-a^2{c}^2\=a^2{x}^2-c^2{x}^2 \+ a^2{y}^{2 }$

$a^2\left(a^2-c^2\right)\=\left(a^2-c^2\right)x^2\+a^2{y}^2$

Now, since the foci lie further from the center than the vertices, $c \> a$, and so $c^2 \> a^2$. We multiply by $-1$ to make both sides positive:

$a^2\left(c^2-a^2\right)\=\left(c^2-a^2\right)x^2-a^2{y}^2$ 

Note that $c^2 \= a^2 \+ b^2$, so $b^2 \= c^2 - a^2$. Substituting $b^2$, we get:

$a^2{b}^2 \= b^2{x}^2- a^2{y}^2$

$1\=\frac{x^2}{a^2}-\frac{y^2}{b^2}$

This is the standard equation for a hyperbola centered at $\left(0\, 0\right)$ with semi-major axis length a and semi-minor axis length b. 

Further from the center of the hyperbola, the branches approach, but never cross, two lines known as asymptotes. For a hyperbola with horizontal major axis, the general equation for these asymptotes is: $y \= \pm \frac{b}{a}\left(x-h\right) \+ k$,

whereas if the major axis is vertical, the general equation for these asymptotes is: $y \= \pm \frac{a}{b}\left(x - h\right) \+ k$.

To prove that these equations represent the asymptotes of a hyperbola, we can look at the most basic case - a hyperbola with horizontal major axis centered at $\left(0\, 0\right)$ with semi-major axis length a and semi-minor axis length b - and can then extend this result to hyperbolae centered away from the origin.

We must prove that the line $y \= \frac{b}{a }x$ is an asymptote of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}\=1$ in the first quadrant (all other quadrants follow by symmetry):

$\frac{x^2}{a^2}-\frac{y^2}{b^2}\=1$

$\frac{x^2}{a^2}-1\=\frac{y^2}{b^2}$

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

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

So, $y \= \sqrt{b^2\left(\frac{x^2}{a^2}-1\right)}$ is the equation of the hyperbola in the first quadrant.

Now, if $y \= \frac{b}{a }x$ is an asymptote of the hyperbola as we claim it to be, this should be the line that the hyperbola approaches as x grows infinitely large.

We can take the limit of the difference between these functions as $x \to \infty$, and if this limit equals 0, then we can be certain that this line is an asymptote.

$L \= \underset{x\to \infty }{lim}\left[\sqrt{b^2\left(\frac{x^2}{a^2}-1\right)} - \frac{b}{a}x\right]$

$L \= \underset{x\to \infty }{lim}\left[\frac{b}{a}\left(\sqrt{x^2 - a^2} - x\right)\right]$

$L \= \underset{x\to \infty }{lim}\left[\frac{b}{a}\frac{\left(x^2 - a^2 - x^2\right)}{\left(\sqrt{x^2 - a^2} \+ x\right)}\right]$$$

$L \= \underset{x\to \infty }{lim}\left[\frac{b}{a}\frac{ -a^2 }{\left(\sqrt{x^2 - a^2} \+ x\right)}\right]$

$L \= -\mathrm{ab}\left(\underset{x\to \infty }{lim}\left[\frac{1}{\left(\sqrt{x^2 - a^2} \+ x\right)}\right]\right)$

$L \= -\mathrm{ab}\left(0\right)$

$L \= 0$

Therefore the branch of the hyperbola, $y \= \sqrt{b^2\left(\frac{x^2}{a^2}-1\right)}$ approaches $y \= \frac{b}{a }x$ as $x \to \infty$ in the first quadrant.