Question

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

Answer 1

it represents the equation of a hyperbola

Explanation:

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

`x=rcostheta and y=rsintheta->r=sqrt(x^2+y^2)`

The given equation

`r=2/(1+2costheta)`

`=>r+2rcostheta=2`

`=>sqrt(x^2+y^2)+2x=2`

`=>(sqrt(x^2+y^2))^2=(2-2x)^2`

`=>x^2+y^2=4-8x+4x^2`

`=>4x^2-x^2-8x-y^2+4=0`

`=>3x^2-y^2-8x+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.

Answer 2

Rearranging, the form is`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

`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 and l =a (e^2-1)`

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

`2/r=1+2 cos theta` and it is revealed that `e = 2 > 1`, and so,

`l = a(e^2-1)=3a=2`. So, a =2/3.

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