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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 $$y= frac{x^2}{4 cdot left(frac{1}{2}pright)}=frac{x^2}{2p}?$$

I thought this because the vertex is halfway between the directrix and the focus of a parabola.

Then I tried to use the facts:
$$r^2 = x^2 +y^2 \
x =rcostheta \
y=rsintheta.$$

But I couldn't get the form required, any corrections, or hints?

Cheers.

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 $y=-p$.

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

$$
frac{cos^2theta}{2p} r^2 - (sin theta) r - frac{p}{2} = 0
$$

If you solve this quadratic expression for $r$, and use the identity $sin^2theta + cos^2theta = 1$ twice, you should obtain the expression you want.

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.

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 $y=r sintheta$

Therefore (1) becomes $$sqrt{r^2}=r sintheta + p$$
$$r = r sintheta + p$$
$$r – r sintheta = p$$
$$r(1 – sintheta) = p$$
Hence $$r = frac{p}{1 – sintheta}$$