Question

椭圆x^2/9+y^2/4=1,过点M(1,1)的直线交双曲线于A,B两点,且M是线段AB中点,求直线AB方程拜托各位了 3Q

Answer 1

设A(x1,y1)B(x2,y2)因为M (1,1)为AB中点所以x1+x2=2 y1+y2=2代入点M设AB直线方程为a(x-1)+1=y 联立方程x1^2/9+y1^2/4=1 x2^2+y2^2=1得4(x1-x2)(x1+x2)=9(y2-y1)(y1+y2)由已知条件可知x1+x2=2 y1+y2=2所以(y1-y2)/(x1-x2)=-4/9即直线斜率为-4/9所以直线方程为y=-4/9x+13/9

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