直线l过点M(1
直线l过点M(1,1)与椭圆x^2/4+y^2/3=1相交于A B两点,若AB的中点为M,求l的方程请写出详细步骤和答案
设A为(x1,y1)、B(x2、y2);分别以点A、B代入椭圆得两式,再两式相减得(x1+x2)(x1-x2)/4+(y1+y2)(y1-y2)/3=0 ==> (y1-y2)/(x1-x2)+3/4*[(x1+x2)/(y1+y2)].而M(1,1)为AB中点,即(x1+x2)/2=1,(y1+y2)/2=1.故k+3/4=0,k=-3/4.故l为y-1=-3/4*(x-1)即3x+4y-7=0
答:直线L:y=x+k, P(x1,y1), Q(x2,y2), M(m,n) ==> y1=x1+k, y2=x2+k, k=n-m ...(1) y=x+k 代...详情>>
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