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

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Answer 1 · 2016-07-30T10:33:50

it represents the equation of a hyperbola

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 · 2016-07-30T12:42:52

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.

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..