User:Wideshanks

Draw a generic parabola and draw a triangle, where $h\,\!$ is the distance from the origin to the focus.

The two legs are $x\,\!$ and $f(x)-h\,\!$, and the hypotenuse is $f(x)+h\,\!$

$x^{2}+(f(x)-h)^{2}=(f(x)+h)^{2}\,\!$

$f(x)^{2}-2c\cdot f(x)+c^{2}+x^{2}=f(x)^{2}+2c\cdot f(x)+c^{2}\,\!$

$-4h\cdot f(x)=-x^{2}\,\!$

$f(x)={\frac {x^{2}}{4h}}\,\!$

In a parabola of the form $f(x)=a\cdot x^{2}\,\!$, the focal distance is ${\frac {1}{4a}}\,\!$.

For $f(x)=x^{2}\,\!$, the focal distance is ${\frac {1}{4}}\,\!$.

$f(x)={\frac {x^{2}}{4{\frac {1}{4}}}}=x^{2}\,\!$