曲线在(a,y(a))处的切线方程为y=y'(a)(x-a)+y(a)它与原点的距离=|-ay'(a)+y(a)|/√(1+y'(a)²)=|a|平方,得：a²y'²+y²-2ayy'=a²(1+y'²)2ayy'=y²-a²将a换成x,则得：2xyy'=y
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