How do you identify the conic of ##r = 3/(1 - sinx)?

Referred to the/a focus S as r =0 and the perpendicular from S to

the ( or corresponding ) directrix as the the initial line #theta = 0#,

the equation to a conic is

(#theta = pi/2# semi-chord length )/r

= 1 + (eccentricity of the conic ) #cos theta# or simply

# l/r = 1+ e cos theta#.

If the perpendicular is along #theta = alpha#, the equation

becomes

#l / r = 1+ e cos ( theta + alpha )#.

The conic is an ellipse, parabola or hyperbola according as

#e < = > 1#

Here,

#3/r = 1 - sin theta = 1 + (1 ) cos ( theta + pi/2 )#

#e = 1#. So, the conic is a parabola, with focus at S ( r = 0 ) and the

perpendicular from S to the directrix is along

#theta = alpha = pi/2#

The Cartesian equivalent is

#3 = r - r sin theta = sqrt( x^2 + y^2 ) - y#

The graph is immediate.
graph{3 - sqrt ( x^2 + y^2) + y = 0}

.

Let’s see by first putting it in rectangular form:

#r(1-sinx)= 3#

#r-rsinx=3#

#r-y=3#

#r=y+3#

Square both sides:

#r^2= y^2+6y+9#

#x^2+y^2= y^2+6y+9#

#x^2= 6y+9#

#(x^2-9)/6= y#

#y= 1/6x^2-3/2#

The polar equation of conic is #r= (ed)/(1-esin (x)# when

directrix is below the pole.

# r= 3 /(1-sin x) :. e=1 ,d=3# If #e = 1#, then the conic is

a parabola. If #e < 1#, then the conic is an ellipse.

If #e > 1=, then the conic is a hyperbola.

Since #e=1# this is parabola , directrix is #3# unit below

focus (pole) and parallel to the polar axis. [Ans]