How do you find the points of intersection of #r=1/sin(Theta)# and #r=3-2sin(Theta)# ?

#r=3-2sin(Theta)#

1 Answer
Nirvi Vakharia
May 17, 2018

Points of intersection in the first quadrant on the polar co-ordinates are:
#(2, pi/6)# and #(1,pi/2)#

Explanation:

The simplest way of finding the point of intersection is to substitute the value of #theta# from the first equation into the second equation. Here, as #theta# is in terms of its sine, we will substitute the value of #sin theta# from the first equation into the second one.

we know,
#r=1/sintheta#
#:, sintheta=1/r# . . . . . . . . [i]

Equation 2:
#r=3-2sintheta# . . . . . . . . . [ii]

Now, at point of intersection, the 2 equations will have the same value of r and #theta#

#:.# Substituting value of #sintheta# from eq.[i] into [ii],

#r=3-2(1/r)#

#:. r^2 = 3r -2# . . . . . . .[ Multiplying both sides by r]
#:. r^2 -3r +2=0#

Now, this is a quadratic equation, which can be factorized as:
#r^2 -2r-r+2=0#

#:. r(r-2)-(r-2)=0#

#:. (r-1)(r-2)=0#

which means,
#r-1=0#
or
#r-2=0#

#:. r=1# or
#r=2#

if r=1,
from eq. [i],

#sintheta=1/1# i.e. #theta = pi/2#

if r=2,
from eq. [i],

#sintheta=1/2# i.e. #theta = pi/6#

Thus, in the first quadrant, the given curves intersect at 2 points,
#(2, pi/6)# and #(1,pi/2)#

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