Question

已知椭圆x^2/a^2+y^2/b^2=1(a>b>0)的离心率是根6/3,过椭圆上一点M作直线MA,MB分别交椭圆于A

Answer 1

∵点A、B关于原点对称,∴可设A(x1,y1),B(-x1,-y1),M(x2,y2)

∴k1=(y1-y2)/(x1-x2),k2=(y1+y2)/(x1+x2)

∴k1*k2=((y1-y2)/(x1-x2))*((y1+y2)/(x2+x1))=(y2^2-y1^2)/(x2^2-x1^2)

又∵A、M在椭圆上,∴x1^2/a^2+y1^2/b^2=1,x2^2/a^2+y2^2/b^2=1

∴两式相减即可：(x1^2-x2^2)/a^2+(y1^2-y2^2)/b^2=0

∴(y2^2-y1^2)/(x2^2-x1^2)=-b^2/a^2=-(a^2-c^2)/a^2=-1+e^2=-1+2/3=-1/3

∴k1·k2=-1/3