直线y=ax+1与椭圆3x^2+y^2=2相交与P、Q两点.当a为何值时以PQ为直径的圆过坐标原点!
问题描述：
直线y=ax+1与椭圆3x^2+y^2=2相交与P、Q两点.当a为何值时以PQ为直径的圆过坐标原点!
答
P(x1,y1)、Q(x2,y2)
联立直线与椭圆, (3+a^2)x^2+2ax-1=0.
韦达定理, x1+x2=-2a/(3+a^2),x1x2=-1/(3+a^2). ----(1)
并且 y1=ax1+1,y2=ax2+1. ----(2)
圆心(x0,y0),半径r: (x-x0)^2+(y-y0)^2=r^2.
以PQ为直径, x1+x2=2x0,y1+y2=2y0,(2r)^2=(x2-x1)^2+(y2-y1)^2. ----(3)
圆过原点, x0^2+y0^2=r^2. ----(4)
联立(1)(2)(3)(4), 得a=1或-1.