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?

  • $\begingroup$ You have $(x-h)$ where you need $(x-h)^2$. $\endgroup$ – Michael Hardy Apr 22 '13 at 22:56

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.

  • $\begingroup$ get my vote in before end of the day :-) $\endgroup$ – robjohn Apr 22 '13 at 23:55

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.

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.

  • $\begingroup$ This answer is very nice, but terrible. One need not know any calculus to learn an answer to this question. Under what circumstances is the second derivative equal to the curvature? Why is the focal length half the radius of curvature? One who knows nothing about those matters can understand an answer to this question, but cannot understand this answer. $\endgroup$ – Michael Hardy Apr 22 '13 at 22:55
  • $\begingroup$ @MichaelHardy: ah, I missed the algebra-precalculus tag. Since you've taken care of the answer from that standpoint, there is not much else for me to do. I don't know if I can come up with anything better than to apply the definition and compute distances. $\endgroup$ – robjohn Apr 22 '13 at 23:21

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