Question

进行直角坐标方程与极坐标方程的互化.

(1)y²=4x;(2)y²+x²-2x-1=0;(3)θ=;(4)ρcos²=1;(5)ρ²cos2θ=4;(6)ρ=.

Answer 1

解:(1)将x=ρcosθ,y=ρsinθ代入y²=4x,得(ρsinθ)²=4ρcosθ,

化简得ρsin²θ=4cosθ.

(2)将x=ρcosθ,y=ρsinθ代入y²+x²-2x-1=0,得(ρsinθ)²+(ρcosθ)²-2ρcosθ-1=0,?

化简得ρ²-2ρcosθ-1=0.?

(3)tanθ=,∴tan==,化简得y=x(x≥0).

(4)∵ρcos² =1,?

∴ρ=1,即ρ+ρcosθ=2.?

∴+x=2,化简得y²=-4(x-1).?

(5)∵ρ²cos2θ=4,∴ρ²cos²θ-ρ²sin²θ=4,即x²-y²=4.?

(6)∵ρ=,

∴2ρ-ρcosθ=1.

∴2,化简得3x²+4y²-2x-1=0.