Proof for the focus of a parabola/parebola :3

In this note I'll prove that the distance of the focus from the vertex (p)(p) of a parabola has the equation 4p(yk)=(xh)24p(y-k)=(x-h)^2 where the parabola has the vertex (h,k).

I would also love feed back on this note. If you look at most of my other notes, I go very far into detail about each step I do to avoid any confusion and to make the note understandable to as many people as possible. However, for the more skilled reader, it may seem a bit long, so this note I'll skip a few simple explanations (no details will be skipped). Tell me what you think in the comments below :) thanks!

Begin with the standard equation of a parabola y=ax2+bx+cy=ax^2+bx+c.

Finding the vertex coordinates using x=b2ax=\frac{-b}{2a}

y=(b24a)b22a+cy=\left(\frac{b^2}{4a}\right)-\frac{b^2}{2a}+c

y=b24a+cy=\dfrac{-b^2}{4a}+c

Our vertex has coordinates

(b2a,b24a+c)\left(\dfrac{-b}{2a},\dfrac{-b^2}{4a}+c\right)

Thus h=b2ah=\dfrac{-b}{2a} and k=b24a+ck=\dfrac{-b^2}{4a}+c

A parabola is the set of points equidistant from the focus and directrix. Thus there is a point (call it m) distance 2p from the focus and distance 2p from the directrix. Therefore, if we draw a line parallel to the directrix from the focus, it will perpendicularly bisect the line from the directrix to m. (Can someone explain why, I can't explain this for some reason, I know why, but I can't put it into words). This is what is depicted in the picture above.

Now, this means there exists a point on our graph distance 2p to the right of our vertex and distance p above. Thus the point has the coordinates (b2a+2p,b24a+c+p)\left(\dfrac{-b}{2a}+2p,\dfrac{-b^2}{4a}+c+p\right). Plugging in for x and y

(b24a+c+p)=a(b2a+2p)2+b(b2a+2p)+c\left(\dfrac{-b^2}{4a}+c+p\right)=a\left(\dfrac{-b}{2a}+2p\right)^2+b\left(\dfrac{-b}{2a}+2p\right)+c

Bash bash bash

1a=4p\dfrac{1}{a}=4p

Going back to our first equation (1a)y=x2+(1a)bx+(1a)c\left(\frac{1}{a}\right)y=x^2+\left(\frac{1}{a}\right)bx+\left(\frac{1}{a}\right)c

4py4pc=x2+4pbx4py-4pc=x^2+4pbx

This part is a little tricky so I'll show all the steps. Add 4p2b24p^2b^2 to both sides.

4p(yc+b2p)=x2+4pbx+4p2b24p(y-c+b^2p)=x^2+4pbx+4p^2b^2

Resubstituting p=14ap=\dfrac{1}{4a} EVERYWHERE BUT THE FIRST P.

4p(yc+b24a)=x2+4pbx+4b216a24p\left(y-c+\frac{b^2}{4a}\right)=x^2+4pbx+\frac{4b^2}{16a^2}

4p(y(b24a+c))=(x(b2a))24p\left(y-\left(\frac{-b^2}{4a}+c\right)\right)=\left(x-\left(\frac{-b}{2a}\right)\right)^2

Remember, h=b2ah=\dfrac{-b}{2a} and k=b24a+ck=\dfrac{-b^2}{4a}+c

4p(yk)=(xh)24p(y-k)=(x-h)^2

And we are done.

Remember, please leave feed back on how you liked/disliked the writing style/comprehensiveness of this note.

Note by Trevor Arashiro
4 years, 9 months ago

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