|
|
Last edited by hbghlyj 2023-2-25 00:04O为原点,OA为x轴正方向,OA=1,以OA为直径作圆,过点A作其切线.作一条弦OBC与圆交于B,与切线交于C.在线段OC上取OM=BC,则
⑴M的轨迹为$ \left(x^{2}+y^{2}\right) x-y^{2}=0 $.
⑵沿y轴取一个长度OD=2,设AD与该曲线交于P,而OP与圆在点A的切线交于Q.证明AQ=$\sqrt[3] 2$.
[上图来自英文维基百科]
证明:
⑴$M(x,y),\frac{y}{\sqrt{x^2+y^2}}=\sin{\angle AOM}=\sin{\angle BAC}={BC\over CA}={OM\over y⋅\frac {OA}x}=\frac{\sqrt{x^2+y^2}}{y⋅\frac 1x}$
⑵设R为QO与圆的交点,AQ²:1=AQ²:AO²=RQ:RO=PO:PQ=OD:AQ=2:AQ.
|
|
