证明曲线{x=a(cost+tsint),y=a(sint-tcost)}上任一点的法线到原点距恒等于a

Question

证明曲线{x=a(cost+tsint),y=a(sint-tcost)}上任一点的法线到原点距
恒等于a

Answer 1

x=a(cost+tsint),y=a(sint-tcost),
∴dx/dt=a(-sint+sint+tcost)=atcost,
dy/dt=a(cost-cost+tsint)=atsint,
∴法线斜率-（dx/dt)/(dy/dt)=-cost/sint,
∴法线方程是xcost+ysint=a(cost+tsint)cost+a(sint-tcost）sint,
即xcost+ysint=a,
∴原点到此法线的距离恒等于a.