求由方程x^2+xy+y^2=4所确定的曲线y=y(x)在点(2,-2)处的切线方程

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求由方程x^2+xy+y^2=4所确定的曲线y=y(x)在点(2,-2)处的切线方程求由方程x^2+xy+y^2=4所确定的曲线y=y(x)在点(2,-2)处的切线方程求由方程x^2+xy+y^2=4

求由方程x^2+xy+y^2=4所确定的曲线y=y(x)在点(2,-2)处的切线方程
求由方程x^2+xy+y^2=4所确定的曲线y=y(x)在点(2,-2)处的切线方程

求由方程x^2+xy+y^2=4所确定的曲线y=y(x)在点(2,-2)处的切线方程
x^2+xy+y^2=4求导得:
2x+y+xy'+2yy'=0,点(2,-2)代入得:
4-2+2y'-4y'=0 y'=1
y(x)在点(2,-2)处的切线方程为y+2=x-2