1.AB垂直x轴时易验证不满足题意,设y=kx,代入椭圆得坐标,得x^2=180/(9k^2+4),y^2=180k^2/(9k^2+4),则│AB│=2√(x^2+y^2)=2[√(k^2+1)]*180/(9k^2+4)F1到直线的距离为d=│-5*k+0*(-1)│/√(k^2+1)S=1/2*d*│AB│,代入得k=±4/3所以直线方程为y=±4/3x 2.点差法设M(X1,y1).N(x2,y2),为直线与双曲线的交点,将点坐标代入曲线方程,得3x1^2-y1^2=33x2^2-y2^2=3两式相减得3(x1^2-x2^2)=y1^2-y2^2变形得(y1-y2)/(x1-x...
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