Derivation of the focus

To derive the focus of a simple parabola, where the axis of symmetry is the y-axis with the vertex at (0,0), such as

$y = a x^2\,\!$

then there is a point (0,f)—the focus, F—such that any point P on the parabola will be equidistant from both the focus and the linear directrix, L. The linear directrix is a line perpendicular to the axis of symmetry of the parabola (in this case parallel to the x axis) and passes through the point (0,-f). So any point P=(x,y) on the parabola will be equidistant both to (0,f) and (x,-f).
FP, a line from the focus to a point on the parabola, has the same length as QP, a line drawn from that point on the parabola perpendicular to the linear directrix, intersecting at point Q.
Imagine a right triangle with two legs, x and f-y (the vertical distance between F and P). The length of the hypotenuse, FP, is given by

$\| FP \| = \sqrt{ x^2 + (f - y)^2 }\,\!$

(Note that (f-y) and (y-f) produce the same result because it is squared.)
The line QP is given by adding y (the vertical distance between the point P and the x-axis) and f (the vertical distance between the x-axis and the linear directrix).

$\| QP \| = f + y\,\!$

These two line segments are equal, and, as indicated above, y=ax², thus

$\| FP \| = \| QP \| \,\!$

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

$x^2 + (f^2 - 2 a x^2 f + a^2 x^4) = (f^2 + 2 a x^2 f + a^2 x^4)\,\!$

Cancel out terms from both sides,

$x^2 - 2 a x^2 f = 2 a x^2 f\,\!$

$x^2 = 4 a x^2 f\,\!$

Divide out the x² from both sides (we assume that x is not zero),

$1 = 4 a f\,\!$

$f = {1 \over 4 a }\,\!$

So, the parabola $\text{[math]}$ can be written as $4fy = x^2\,\!$ ; and for a parabola such as f(x)=x², the a coefficient is 1, so the focus F is (0,¼)

As stated above, this is the derivation of the focus for a simple parabola, one centered at the origin and with symmetry around the y-axis. For any generalized parabola, with its equation given in the standard form

$\text{[math]}$ ,