设曲线C的极坐标方程多p＾2十2Pcos(a十兀/4)一4=0直线l的参数方程为{x=1+2分之根号2t y=-2-2分

1) 由x=pcosa,y=psina,p²=x²+y²得
p²十2Pcos(a十π/4)一4=0
p²十2P[(√2/2)cosa-(√2/2)sina]一4=0
p²十√2pcosa-√2psina]一4=0
x²+y²+√2x-√2y-4=0
(x+√2/2)²+(y-√2/2)²=5
2)因为 参数方程为{x=1+2分之根号2t y=-2-2分之根号2t
是以位移为参数的参数方程,所以把x=1+2分之根号2t ,y=-2-2分之根号2t代入
(1+√2t/2+√2/2)²+(-2-√2t/2-√2/2)²=5
(1+t)²+3√2(1+t)=0
∴t1=-1,t2=-1-3√2
线段长|AB|=|t1-t2|=|-1-3√2+1|=3√2
作参考吧

解1由p＾2十2Pcos(a十兀/4)一4=0
即p＾2十2Pcosacosπ/4-2psinasinπ/4一4=0
即p²=x²+y² pcosa=x，psina=y
得x²+y²+√2x-√2y-4=0
即（x+√2/2)²+（y-√2/2)²=5
即曲线C的极坐标方程化为直角方程（...

57 分钟前
1) 由x=pcosa, y=psina, p²=x²+y²得
p²十2Pcos(a十π/4)一4=0
p²十2P[(√2/2)cosa-(√2/2)sina]一4=0
p²十√2pcosa-√2psina]一4=0
x²+y²+√2x-√2y-4=0
(x...

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p²+2pcos(a+π/4)+4=0
p²+2pcosα*cos(π/4)-2psinα*sin(π/4)-4=0
p²+√2pcosα-√2psinα-4=0
p²＝x²+y²，pcosα＝x，psinα＝y
x²+y²+√2x－√2y－4＝0
(x+√2/2...