Is there an elegant way to derive the quadratic equation from the definition of a parabola?
In other words, how can it be proven that the set of all points an equal distance away from a focus and a diretrix forms a curve that can be represented by the equation y=ax^2 ?
I'm sure that this is easy to do using calculus, but I'm wondering if a purely algebraic/geometric proof is possible.
Finding the formula for a parabola in standard position doesn't need calculus:
Position your coordinate system so that the focus is at the point $(0,h)$ where $h\gt 0,$ and the directrix is the line $y=-h.$
Then $(x,y)$ is a point on the parabola iff the distance from $(x,y)$ to $(0,h)$ equals the distance from $(x,y)$ to the line $y=-h,$ and the distance from $(x,y)$ to the line is the same as the distance from $(x,y)$ to $(x,-h),$ which is just $\lvert y+h\rvert.$ It follows that $(x,y)$ is on the parabola iff
$$\sqrt{x^2+(y-h)^2}=\lvert y+h\rvert.$$
Now, applying the facts that both sides of the above equation are non-negative and that $(y+h)^2=\lvert y+h\rvert^2,$ we have that the point $(x,y)$ is on the parabola iff
$$x^2+(y-h)^2=(y+h)^2.$$
Multiplying these out yields
$$x^2+y^2-2hy+h^2=y^2+2hy+h^2,$$
and canceling out the $y^2$ and the $h^2$ gives us
$$x^2=4hy,$$
or
$$y=\frac1{4h}x^2,$$
as desired.
Edit: Thanks to @egreg for a suggestion on improving the presentation of the answer.