过曲线x^2+xy+y^2=4上点(2,-2)处的切线方程是?

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过曲线x^2+xy+y^2=4上点(2,-2)处的切线方程是?过曲线x^2+xy+y^2=4上点(2,-2)处的切线方程是?过曲线x^2+xy+y^2=4上点(2,-2)处的切线方程是?都对x求导得:

过曲线x^2+xy+y^2=4上点(2,-2)处的切线方程是?
过曲线x^2+xy+y^2=4上点(2,-2)处的切线方程是?

过曲线x^2+xy+y^2=4上点(2,-2)处的切线方程是?
都对x求导得:2x+y+xy‘+2yy'=0
把(2,-2)这一点代进去:得y'=1
则切线方程是:y=x-4

∵x²+xy+y²=4 ==>2x+y+xy'+2yy'=0 (对x求导数)
==>y'=-(2x+y)/(x+2y)
∴在点(2,-2)处的斜率k=-(2*2-2)/(2-2*2)=-1
∵切线过点(2,-2)
∴切线方程是y=-(x-2)-2,即y=-x。