Mathematics, 17.07.2021 08:40 weeblordd

Where does the 4p come from in (x-a)^2 = 4p(y-k) when dealing with the focus and directrix of a parabola. Going to use (a, b) as the focus and y = k for the directrix

Mostly going to be talking about vertical parabolas, but I think I can use whatever answers I get to figure out horizontal ones. Everything I'm writing out here I've figured out for both.

Equating the distance between the focus and a point on the graph (namely the vertex) and the vertex and directrix gets me $\sqrt{(y-k)^2}=\sqrt{(x-a)^2+(y-b)^2}$

Then expanding and rearranging I get $y=\frac{1}{2(b-k)}(x-a)^2 +.5(b+k)$ Whichh I rearrange to because it's in y=f(x) form and it's basically a complicated vertex form.

Using that information and assuming I have a simple vertex form of a parabola and want to find the focus and directrix, where my Parabola is in the form A(x-C) + D $\frac{1}{2A} = b - k$ and 2D = b + k, which is enough to find the focus and directrix since the x coordinate of the focus will be a in $y=\frac{1}{2(b-k)}(x-a)^2 +.5(b+k)$ or C in A(x-C)^2 + D (for vertical parabolas anyway)

I've seen the whole (x-a)^2 = 4p(y-k) form though and I'm wondering where that is from, and how it is derived. According to it the focus would be (h, k+p) and the directrix would be y = k-p. Totally expecting it to be simple and something I'm just missing lol

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Where does the 4p come from in (x-a)^2 = 4p(y-k) when dealing with the focus and directrix of a para...

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