# Finding equation of parabola

Find an equation of the parabola with focus at point $(0,5)$ whose directrix is the line $y=0$. (Derive this equation using the definition of the parabola as a set of points that are equidistant from the directrix and the focus)

Ok this one is killing me. My textbook has this

An equation of the parabola with focus $(0,p)$ and directrix $y=-p$ is $x^2=4py$

However, my problem has $(0,p)$ with p being 5, yet the directrix is not $p=-y=-5$, it's $0$. And so I tried deriving it manually as they said by

$\sqrt{x^2+(y-p)^2}=|y+p|$

But that just takes me back to the original equation I mentioned, $x^2=4py$. Not sure what I'm supposed to do.

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Edited to try to clarify OP's confusion:

The distance between $(x,y)$ and $(0,5)$ is indeed $\sqrt{x^2+(y-5)^2}$.

The distance between $(x,y)$ and the line $y=0$ is $|y|$ (if $y>0$, then it is above the $X$-axis, and the distance is just the $y$ coordinate; if $y<0$, then the distance is $-y = |y|$). So the points on the parabola are exactly the points for which the two distances are equal, that is, all $(x,y)$ for which: $$\sqrt{x^2+(y-5)^2} = |y|.$$ Now square both sides, and you'll get the equation of the parabola you want.

How is this related to the formula you quote? It's really all just the same process. If the directrix is either horizontal (a line of the form $y=k$) or vertical (a line of the form $x=\ell$), then it is very easy to compute the distance from a point to the directrix: the point $(x,y)$ is $|y-k|$ away from the line $y=k$, and is $|x-\ell|$ away from the line $x=\ell$. If the focus of the parabola is at $(a,b)$, then you want the distance from $(x,y)$ to $(a,b)$ to equal the distance to the directrix, so you get: \begin{align*} \sqrt{(x-a)^2 + (y-b)^2} &= |y-k| &\qquad&\mbox{if the directrix is $y=k$,}\\ \sqrt{(x-a)^2 + (y-b)^2} &= |x-\ell| &&\mbox{if the directrix is $x=\ell$.} \end{align*} Then squaring both sides yields the equation of the parabola. Doing it correctly will cancel out the $y^2$ term when the directrix is horizontal, and the $x^2$ term when the directrix is vertical.

For more general parabolas, when the directrix is neither horizontal nor vertical but an arbitrary line $y=mx+b$, you need to work a bit harder, because the distance from $y=mx+b$ to $(x,y)$ is not so simple to compute. Not too hard, but not as simple. Once you have a formula for the distance, you set it equal to the distance to the focus, square both sides, and get the equation of the parabola.

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Ok so I get $x^2=2py-p^2$. However, I'm confused between whats p and whats y. Whenever we mentioned y so far, has it been about the same y, the one thats equal to 0? Because then we have the point (x,y)=(0,5) in which y=5, so thats confusing me a bit. – maq Dec 8 '10 at 22:14
Ok wait I'm not making any sense. I just realized that p is just a general variable they're using. In our case, p=0, so wouldnt it just be $\sqrt{x^2+y^2}=5$? But then thats a circle.. – maq Dec 8 '10 at 22:22
In Arturo Magidin's answer, $p=5$. The $p$ in your question doesn't work here, because you need the directrix to be $y=-p$. But, you have $y=0$ and $p=5$. – Joe Johnson 126 Dec 8 '10 at 22:43
@f-Prime: I've rewritten, hopefully that will clarify. – Arturo Magidin Dec 9 '10 at 2:59

So, you have a method for writing an equation for a parabola where the focus is at $(0,p)$ and the directrix is $y=-p$, but now you want to write an equation for a parabola with focus $(0,5)$ and directrix $y=0$, which doesn't fit the formula.

If you take the conditions you are given and translate them down 2.5 units, the focus would be at $(0,2.5)$ and the directrix would be at $y=-2.5$, so you should be able to write an equation for that parabola. Now, translate that equation up 2.5 units (replace $y$ with $(y-2.5)$).

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From wikipedia:

Equation for a general parabola with a focus point $F =(u, v)$, and a directrix in the form

$$n_1x+n_2y+c=0 \,$$ is

$$\frac{\left|n_1x+n_2y+c\right|}{\sqrt{{n_1}^{2}+{n_2}^{2}}}=\sqrt{\left(x-u\right)^2+\left(y-v\right)^2}$$

So make the substitutions and verify that the what you get agrees with Arturo Magidin's derivation.

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