What is the rationale for the factor of $4$ in the Conics parabola equation?

Question

The Conics form of a parabola equation is $4p(y-k)=(x-h)^2$ where $(h,k)$ is the vertex of the parabola and $p$ is the distance from the vertex to the focus. (Which is also the same distance from the vertex to the directrix). My question is where does this factor of $4$ come from?

Answer 1

Precalculus Answer

Let's translate so that the vertex is $(h,k)=(0,0)$ and compute the locus of points equidistant from the directrix $y=-p$ and the focus $(0,p)$. $$ \begin{align} x^2+(y-p)^2&=(y+p)^2\\ x^2+y^2-2py+p^2&=y^2+2py+p^2\\ x^2&=4py \end{align} $$

Therefore, $4$ is there so that $p$ is the distance from the vertex of the parabola to the focus.

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Calculus Answer

Taking a couple of derivatives, we have $$ \begin{align} 4p(y-k)&=(x-h)^2\\ 4py'&=2(x-h)\\ 4py''&=2\\ 2py''&=1 \end{align} $$ Since $y''$ at $(h,k)$ is the curvature of the parabola at the vertex, $2p$ is the radius of curvature, and therefore, $p$ is the focal length.

Therefore, $4$ is there so that $p$ is the focal length.

Answer 2

If the equation were given as $p(y-k)=(x-h)^2$ rather than as $4p(y-k)=(x-h)^2$, then the distance from the vertex to the focus would be $p/4$ rather than $p$. In other words, the "$4$" is there in order that $p$ will be that distance.

Consider the case where the vertex is $(h,k)=(0,0)$. The equation is $4py=x^2$. According to what you say you've read, the focus should be $(0,p)$. Let's check that that is indeed the focus. Remember the basic characterization in terms of focus and directrix: The distance from a point on the curve to the focus is always the same as the distance from that point to the directrix. Since this parabola has a vertical axis (i.e. the $y$-axis, not the $x$-axis or a line going in some other direction), the directrix should then be the line $y=-p$.

The distance from the point $(x,y)=(x,x^2/(4p))$ on the curve to the directrix is then $y+p$. The distance from $(x,y)=(x,x^2/(4p))$ to the focus is the distance from $(x,x^2/(4p))$ to $(0,p)$. That distance is $$ \sqrt{(y-p)^2+(x-0)^2} =\sqrt{(y-p)^2+4py} = \sqrt{(y^2-2py+p^2)+4py} $$ $$ =\sqrt{y^2+2py+p^2} = \sqrt{(y+p)^2} = y+p. $$ So the two distances are indeed equal.