# How do you identify the conic of  r = 2/(1 + 2 cosx)?

Jul 30, 2016

it represents the equation of a hyperbola

#### Explanation:

We know that the cartesian coordinate $\left(x , y\right)$ of a point is related with its polar coordinate $\left(r , \theta\right)$ as follows:

$x = r \cos \theta \mathmr{and} y = r \sin \theta \to r = \sqrt{{x}^{2} + {y}^{2}}$

The given equation

$r = \frac{2}{1 + 2 \cos \theta}$

$\implies r + 2 r \cos \theta = 2$

$\implies \sqrt{{x}^{2} + {y}^{2}} + 2 x = 2$

$\implies {\left(\sqrt{{x}^{2} + {y}^{2}}\right)}^{2} = {\left(2 - 2 x\right)}^{2}$

$\implies {x}^{2} + {y}^{2} = 4 - 8 x + 4 {x}^{2}$

$\implies 4 {x}^{2} - {x}^{2} - 8 x - {y}^{2} + 4 = 0$

$\implies 3 {x}^{2} - {y}^{2} - 8 x + 4 = 0$

This is the cartesian form of the given polar equation.It is obvious from the equation that it represents the equation of a hyperbola.

Jul 30, 2016

Rearranging, the form is$\frac{l}{r} = 1 + e \cos \theta$ that represents a conic. The eccentricity e = 2 > 1. So, the conic is a hyperbola.

#### Explanation:

The polar equation of a conic referred to a focus as pole and the

straight line from the pole to the center of the conic as the initial line

($\theta = 0$) is

$\frac{l}{r} = 1 + e \cos \theta$, where e is the eccentricity and l = semi latus

rectum.

This is derived using the property that

'the distance from the focus = eccentricity X distance from the

(corresponding ) directrix.

For a hyperbola, the eccentricity $e > 1 \mathmr{and} l = a \left({e}^{2} - 1\right)$

Now, the given equation can be rearranged to this standard form

$\frac{2}{r} = 1 + 2 \cos \theta$ and it is revealed that $e = 2 > 1$, and so,

$l = a \left({e}^{2} - 1\right) = 3 a = 2$. So, a =2/3.

The semi transverse axis b = a sqrt(e^2-1)=(2/3)sqrt 3. > a..