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

3 Answers
Aug 9, 2018

Parabola. See explanation.

Explanation:

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}

.

Aug 9, 2018

Parabola

Explanation:

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

Aug 9, 2018

Since e=1 this is parabola , directrix is 3 unit below
focus (pole) and parallel to the polar axis.

Explanation:

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]