Question

设A,B是直线y=2x-3与椭圆x^2/4+y^2=1的两个交点,M是AB的中点,O为坐标原点,则直线OM的斜率是?

Answer 1

y=2x-3代入方程得：x^2/4+(2x-3)^2=1

x^2+4(4x^2-12x+9)=4

17x^2-48x+32=0

x1+x2=48/17

y1+y2=2(x1+x2)-6=-6/17

M的坐标是：x=（x1+x2)/2=24/17,y=(y1+y2)/2=-3/17

那么OM的斜率是K=y/x=-1/8