题目内容

(本题满分15分)

已知椭圆C:+=1(ab>0)的离心率为,且经过点P(1,).

(1)求椭圆C的方程;

(2)设F是椭圆C的右焦点,M为椭圆上一点,以M

圆心,MF为半径作圆M.问点M横坐标满足什么条

件时,圆My轴有两个交点?

(3)设圆My轴交于DE两点,

求点DE距离的最大值.

解:(1)∵椭圆+=1(ab>0)的离心率为,且经过点P(1,),

∴,即 ,解得 ,………………3分

∴椭圆C的方程为+=1。………………5分

(2)易求得F(1,0)。设M(x0y0),则+=1, 圆M的方程为(x-x0)2+(y-y0)2=(1-x0)2+y02

x=0,化简得y2-2y0y+2x0-1=0,⊿=4y02-4(2x0-1)2>0……①。

y02=3(1-)代入①,得3x02+8x0-16<0,解出 -4<x0<故-2≤x0<……9分

(3)设D(0,y1),E(0,y2),其中y1y2。由(2),得

DE= y2- y1===,

x0=-时,DE的最大值为。………………15分

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