Deriving the polar equation of a parabola with focus at the origin and directrix at $y=-p$ 
How is the following result of a parabola with focus $F(0,0)$ and directrix $y=-p$, for $p \gt 0$ reached? It is said to be $$r(\theta)=\frac{p}{1-\sin \theta} $$

I started by saying the the standard equation of a parabola, in Cartesian form is $y= \frac{x^2}{4p} $, where $p \gt 0 $ and the focus is at $F(0,p)$ and the directrix is $y=-p$. So for the question above, would the equation in Cartesian form be $$y= \frac{x^2}{4 \cdot \left(\frac{1}{2}p\right)}=\frac{x^2}{2p}?$$
I thought this because the vertex is halfway between the directrix and the focus of a parabola.
Then I tried to use the facts:
$$r^2 = x^2 +y^2 \\
x =r\cos\theta \\
y=r\sin\theta.$$
But I couldn't get the form required, any corrections, or hints?
Cheers.
 A: For a parabola with $F(0,0)$ and directrix $y=-p$, first write a “distance equation” relating any point on the parabola, $P(x,y)$, to $F$ and to the directrix, $D$.
A parabola can be defined by its locus: $distanceFP = distanceDP$.
$distanceFP = \sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2}$
$distanceDP = y-(-p) = y+p$
So, for $P(x,y)$ on the parabola with $F(0,0)$ and directrix $y=-p$, we can write
$$\sqrt{x^2+y^2}=y+p   …(1)$$
For polar cords, we know that $x^2+y^2=r^2$, and $y=r sin\theta$
Therefore (1) becomes $$\sqrt{r^2}=r sin\theta + p$$
$$r = r sin\theta + p$$
$$r – r sin\theta = p$$
$$r(1 – sin\theta) = p$$
Hence $$r = \frac{p}{1 – sin\theta}$$
A: No, that can't work, because you've shown that these parabolae have different foci.
Since there are lots of parabolae ($x^2=4ay$, $y^2=4ax$, any rotation and translation of one of these), it's easier to work backwards. So let's do that: $\sin{\theta} = y/r$, so
$$ p= r(1-\sin{\theta}) = r-y, $$
so
$$ x^2+y^2 = r^2 = (p+y)^2 = p^2+2py+y^2, $$
and
$$ x^2 = 2py+p^2. $$
In particular, if we look more closely, we notice that
$$ \sqrt{x^2+y^2} = p+y  $$
is the definition of a parabola: the distance from $(0,0)$ to the point is the same as $y+p$, the distance from the horizontal line $y=-p$.
A: The equation of the parabola you want is
$$
y = \frac{x^2}{2p} - \frac{p}{2}
$$
Substituting
$$
x = r \cos \theta
$$
$$
y = r \sin \theta
$$
gives us
$$
\frac{\cos^2\theta}{2p} r^2 - (\sin \theta) r - \frac{p}{2} = 0
$$
If you solve this quadratic expression for $r$, and use the identity $\sin^2\theta + \cos^2\theta = 1$ twice, you should obtain the expression you want.
A: The equation for a parabola in the complex plane is
$$z=\frac{1}{2}p(u+i)^2\\
y=pu\\
x=\frac{1}{2}p(u^2-1)
$$
I think you would have to say
$$r=|z|\\
\theta=\arg(z)$$
to get the true polar form.
Ref: Zwikker, C. (1968). The Advanced Geometry of Plane Curves and Their Applications, Dover Press.
