已知椭圆C方程4x^2+9y^2=36,直线y=kx+m与椭圆C交于AB两点,且以AB为直径的圆恰好过椭圆右顶点

4x²+9(kx+m)²=36(4+9k²)x²+18kmx+(9m²-36)=0由韦达定理：x1+x2=-18km/(4+9k²)x1x2=(9m²-36)/(4+9k²)因为A2A⊥A2B(A(a,0)[(-y1)/(x1-a)][(-y2)/x2-a)]=-1y1y2+x1x2-a(x1+x2)+a²=0y1y2=(kx1+m)(kx2+m)=k²x1x2+m(x1+x2)+m²(1+k²)x1x2++(m-a)(x1+x2)+(a²+m²)=0(1+k²)(9m²-36)/(4+9k²)+(m-a)[-18km/(4+9k²)]+(a²+m²)=0运算量太大待续