Question

椭圆的一个性质怎么证明椭圆x^2/a^2+y^2/b^2=1,过(m,0)(m＞a)作直线l交椭圆于A,B两点,E是B关

Answer 1

证明：设AB直线方程是x=ky+m

A(x1.y1),B(X2,Y2),E(X2,-Y2)

x^2/a^2+y^2/b^2=1

(ky+m)^2b^2+a^2y^2=a^2b^2

(k^2y^2+m^2+2kym)b^2+a^2y^2-a^2b^2=0

(k^2b^2+a^2)y^2+2kmb^2y+m^2b^2-a^2b^2=0

等一下再来答．