X=a(cosθ+cos2θ),y=b(sinθ+sin2θ) Remove θ By applying Elimination Methods ??

1 Answer
P dilip_k
Jun 1, 2018

Given

#x=a(cosθ+cos2θ)....(1)#

#y=b(sinθ+sin2θ).....(2)#

From (1) and (2) we get

#x^2/a^2+y^2/b^2=(sintheta+sin2theta)^2+(costheta+cos2theta)^2#

#=>x^2/a^2+y^2/b^2=(sin^2theta+cos^2theta)+(sin^2 2theta+cos^2 2theta)+2(cos2thetacostheta+sin2thetasintheta)#
#=>x^2/a^2+y^2/b^2=1+1+2cos(2theta-theta)#

#=>x^2/a^2+y^2/b^2=2(1+costheta)....(3)#

Now by (1) we have

#x=a(cosθ+cos2θ)#

#=>x/a=2cos^2θ+costheta-1#

#=>x/a=2cos^2θ+2costheta-costheta-1#

#=>x/a=2cosθ(costheta+1)-(costheta+1)#

#=>x/a=(2cosθ-1)(costheta+1).....(4)#

Combining (3) and (4) we get

#=>x/a=(x^2/a^2+y^2/b^2-2-1)*1/2(x^2/a^2+y^2/b^2)#

#=>2x/a=(x^2/a^2+y^2/b^2-3)(x^2/a^2+y^2/b^2)#

#=>(x^2/a^2+y^2/b^2-3)(x^2/a^2+y^2/b^2)=(2x)/a#

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