四月新
幼苗
共回答了11个问题采纳率:100% 举报
(2设P(x0,y0),A(3,0),M(9/2,yM)
过点P做PB垂直于AF,设右准线与与x轴的交点为N,则PB:MN=FB:FN
即y0/yM=(x0+2)/(9/2+2)
即yM=(13y0/2)/(x0+2)
k1=y0/(x0-3),k2=yM/(9/2-3)
k1·k2=y0/(x0-3)*yM/(9/2-3)
=2y0yM/[3(x0-3)]
=13y0*y0/[3(x0-3)(x0+2)]
x0^2/9+y0^2/5=1,y0^2=5/9(9-x0^2)
k1·k2=(65/27)*(9-x0^2)/[(x0-3)(x0+2)]
=-(65/27)*(x0+3)/(x0+2)
=-(65/27)*[1+1/(x0+2)]
FM交椭圆C于P,-2
1年前
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