How do you convert #r=6 cos θ + 4 sin θ# into rectangular form?
1 Answer
Feb 12, 2016
# x^2 + y^2 - 6x - 4y = 0#
Explanation:
using the formulae that links Polar to Rectangular coordinates.
#• r^2 = x^2 + y^2 #
#• x = rcostheta rArr costheta = x/r #
#• y = rsintheta rArr sintheta = y/r # in the above question then
# r = 6.(x/r) + 4.(y/r) # ( multiplying both sides by r )
# r^2 = 6x + 4y#
# rArr x^2 + y^2 -6x - 4y = 0 # which is the equation of a circle : centre(3,2 ) and
# r =sqrt13 #
