椭圆ax^2+by^2=1与直线y=1-x交与A,B两点,过原点与线段AB中点的直线斜率为二分之根号3,则a/b的值为多少?
问题描述：
椭圆ax^2+by^2=1与直线y=1-x交与A,B两点,过原点与线段AB中点的直线斜率为二分之根号3,则a/b的值为多少?
答
设两点分别为(x1,y1),(x2,y2)分别代入椭圆方程得：ax1^2+by1^2=1ax2^2+by2^2=1两式相减,得：a(x1^2-x2^2)+b(y1^2-y2^2)=0展开得：a(x1+x2)(x1-x2)+b(y1+y2)(y1-y2)=0移项,整理得：(y1-y2)/(x1-x2)= - (a/b)*(x1+x2)/...