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 $$\frac{1}{4}\,\!$$.

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