高二數(shù)學:橢圓問題
來源:學大教育 時間:2013-12-22
知橢圓C:x^2/a^2+y^2/b^2=1的左焦點F,右頂點A,動點M為右準線上一點(異于右準線與x軸的交點),FM交橢圓C于P,已知橢圓C的離心率為2/3,點M的橫坐標為9/2。設(shè)直線PA的斜率為k1,直線MA的斜率為k2,求k1·k2的取值范圍
設(shè)P(x0,y0),A(3,0),M(9/2,yM) 過點P做PB垂直于AF,設(shè)右準線與與x軸的交點為N,
則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交橢圓C于P,-2