Question

已知椭圆C(x^2/a^2)+(y^2/b^2)=1(a>b>0)和直线y=kx+b交于A,B两点求|AB|长

Answer 1

交点为（x1,y1),(x2,y2),联立(a^2*k^2+b^2)x^2+2kba^2x=0,有公式（须牢记）|AB|=sqrt[(1+k^2)(x1-x2)^2]=sqrt[(1+k^2)*(-2kba^2)^2/(a^2*k^2+b^2)^2]=sqrt(1+k^2)*[(2kba^2)/(a^2*k^2+b^2)].