Question

已知曲线C方程为16x^2+4y^2-32xcosθ-16y(sinθ)^2-4(sin2θ)^2=0 (θ属于R),求

Answer 1

16x^2+4y^2-32xcosθ-16y(sinθ)^2-4(sin2θ)^2=

16(x^2-2xcosθ+cos^θ)+4(y^2-4y(sinθ)^2+(sinθ)^4)=4(sin2θ)^2+16cos^θ-4(sinθ)^4

16(x-cosθ)^2+4(y-2(sinθ)^2)=4(sin2θ)^2+16cos^θ-4(sinθ)^4

中心（x,y)

x=-cosθ

y=-2(sinθ)^2

-2(sinθ)^2=-1(1-x^2)=x^2-1

所以y=x^2-1