高中圆锥曲线结论June 6, 2021 · One min read若椭圆(b2>0b^2>0b2>0)或双曲线(b2<0b^2<0b2<0) x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1a2x2+b2y2=1 与直线 y=kx+my=kx+my=kx+m 交于A(x1,y1),B(x2,y2)A(x_1,y_1),B(x_2,y_2)A(x1,y1),B(x2,y2),联立消 yyy (b2+a2k2)x2+2kma2x+a2(m2−b2)=0(b^2+a^2k^2)x^2+2kma^2x+a^2(m^2-b^2)=0(b2+a2k2)x2+2kma2x+a2(m2−b2)=0 得 Δ=4a2b2(b2+a2k2−m2)≥0\Delta = 4a^2b^2(b^2+a^2k^2 - m^2) \geq 0Δ=4a2b2(b2+a2k2−m2)≥0 x1+x2=−2kma2b2+a2k2,x1x2=a2(m2−b2)b2+a2k2x_1+x_2=\frac{-2kma^2}{b^2+a^2k^2}, x_1x_2=\frac{a^2(m^2-b^2)}{b^2+a^2k^2}x1+x2