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 $\mathrm{PE}equals;\sqrt{{\left(x-\left(-c\right)\right)}^{2}plus;{\left(y-0\right)}^{2}}equals;\sqrt{{\left(xplus;c\right)}^{2}plus;{y}^{2}}$.

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

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:

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

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

$\sqrt{{\left(x\+c\right)}^{2}\+{y}^{2}}equals;2aplus;\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}$

${\left(x\+c\right)}^{2}\+{y}^{2}equals;{\left(2aplus;\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}\right)}^{2}$

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

${x}^{2}\+2cx\+{c}^{2}\+{y}^{2}equals;4{a}^{2}plus;4a\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}plus;{x}^{2}-2cxplus;{c}^{2}plus;{y}^{2}$

$2cxequals;4{a}^{2}plus;4a\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}-2cx$

$4cx-4{a}^{2}equals;4a\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}$

$cx-{a}^{2}equals;a\sqrt{{\left(x-c\right)}^{2}plus;{y}^{2}}$

${\left(cx-{a}^{2}\right)}^{2}equals;{a}^{2}\left({\left(x-c\right)}^{2}plus;{y}^{2}\right)$

${c}^{2}{x}^{2}-2{a}^{2}cx\+{a}^{4}equals;{a}^{2}{x}^{2}-2{a}^{2}cxplus;{a}^{2}{c}^{2}plus;{a}^{2}{y}^{2}$

${c}^{2}{x}^{2}\+{a}^{4}equals;{a}^{2}{x}^{2}plus;{a}^{2}{c}^{2}plus;{a}^{2}{y}^{2}$

${a}^{4}-{a}^{2}{c}^{2}equals;{a}^{2}{x}^{2}-{c}^{2}{x}^{2}plus;{a}^{2}{y}^{2}$

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

Now, since the foci lie further from the center than the vertices, $cgt;a$, and so ${c}^{2}gt;{a}^{2}$. 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 ${c}^{2}equals;{a}^{2}plus;{b}^{2}$, so ${b}^{2}equals;{c}^{2}-{a}^{2}$. Substituting ${b}^{2}$, we get:

${a}^{2}{b}^{2}equals;{b}^{2}{x}^{2}-{a}^{2}{y}^{2}$

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

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.

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