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

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

($\theta = \frac{\pi}{2}$ semi-chord length )/r

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

$\frac{l}{r} = 1 + e \cos \theta$.

If the perpendicular is along $\theta = \alpha$, the equation

becomes

$\frac{l}{r} = 1 + e \cos \left(\theta + \alpha\right)$.

The conic is an ellipse, parabola or hyperbola according as

$e < = > 1$

Here,

$\frac{3}{r} = 1 - \sin \theta = 1 + \left(1\right) \cos \left(\theta + \frac{\pi}{2}\right)$

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

perpendicular from S to the directrix is along

$\theta = \alpha = \frac{\pi}{2}$

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:

Letâ€™s see by first putting it in rectangular form:

$r \left(1 - \sin x\right) = 3$

$r - r \sin x = 3$

$r - y = 3$

$r = y + 3$

Square both sides:

${r}^{2} = {y}^{2} + 6 y + 9$

${x}^{2} + {y}^{2} = {y}^{2} + 6 y + 9$

${x}^{2} = 6 y + 9$

$\frac{{x}^{2} - 9}{6} = y$

$y = \frac{1}{6} {x}^{2} - \frac{3}{2}$

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.

$r = \frac{3}{1 - \sin x} \therefore e = 1 , d = 3$ 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]