Question

x^2/12+y^2/3=1,过椭圆c的右焦点于椭圆c分别交于A,B两点,A在x轴下方,且向量AF=3向量FB,F为椭圆右焦点

求过O,A,B三点的圆的方程

Answer 1

椭圆C的方程为x^2/12+y^2/3=1
设过右焦点的直线为x=my+3==>x^2=m^2y^2+6my+9
代入椭圆得(m^2+4)y^2+6my-3=0
设A点坐标为（x1,y1) B点坐标为（x2,y2)
由韦达定理得Y1+y2=-6m/(m^2+4)，y1y2=-3/(m^2+4)
∵向量AF=3向量FB
∴y1+3y2=0
联立解得y1=-9m/(m^2+4),y2=3m/(m^2+4)
∴-27m^2/(m^2+4)^2=-3/(m^2+4)==>m1=-√2/2，m2=√2/2
∵点A在x轴下方，∴取m=√2/2

过右焦点的直线为y=√2x-3√2
∴y1=-√2,y2=√2/3==>x1=2,x2=10/3
∴A(2,-√2)，B(10/3, √2/3)
AB中点坐标为(8/3,-√2/3)
∴AB中垂线方程为y=-√2/2x+√2
OA中点坐标为(1, -√2/2)
∴OA中垂线方程为y=√2x-3√2/2
二中垂线方程联立解得x=5/3，y=√2/6
∴过O,A,B的圆的圆心为(5/3, √2/6)，半径r=√102/6
∴过O,A,B的圆方程为(x-5/3)^2+(y-√2/6)^2=17/6