Question

设直线与椭圆交于（x1，y1）和（x2，y2）两点，求x1·y2+x2·y1

设直线与椭圆方程分别为Ax+By+C=0与x^2/a^2+y^2/b^2=1

Answer 1

由Ax+By+C=0得y=-(Ax+C)/B,①
代入x^2/a^2+y^2/b^2=1得
b^2x^2+a^2(Ax+C)^2/B^2=a^2b^2,
去分母，整理得(b^2B^2+a^2A^2)+2a^2ACx+a^2C^2-a^2b^2B^2=0,
则x1+x2=-2a^2AC/(a^2A^2+b^2B^2),
x1x2=(a^2C^2-a^2b^2B^2)/(a^2A^2+b^2B^2),
由①，x1·y2+x2·y1=x1[-(Ax2+C)/B]+x2[-(Ax1+C)/B]
=-[2Ax1x2+C(x1+x2)]/B
=-[2A(a^2C^2-a^2b^2B^2)-2a^2AC^2]/[B(a^2A^2+b^2B^2)]
=2a^2b^2AB/(a^2A^2+b^2B^2).

Answer 2

简单分析一下，详情如图所示