(2018金华十校高二上期末)已知椭圆$C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)$,右焦点$F_2(2,0)$,点$(\sqrt{3},1)$在椭圆上.
(1)求椭圆的方程;
(2)设$P(x_0,y_0)(y_0>0)$为椭圆$C$上一点,过焦点$F_1,F_2$的弦分别为$PA,PB$,设$\vv{PF_1}=\lambda_1\vv{F_1A},\vv{PF_2}=\lambda_2\vv{F_2B}$,若$\lambda_1=2$,求$\lambda_2$的值.
第一问答案：
$\frac{x^2}{6}+\frac{y^2}{2}=1.$ |
kuing
Posted 2019-6-26 21:54
