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
