第1个回答 2019-01-19
(Ⅰ)抛物线C1:y2=4x的焦点F为(1,0),
由题意可得a2−b2=1①
由C1与C2关于x轴对称,可得C1与C2的公共点为(23,±26√3),
可得49a2+83b2=1②
由①②解得a=2,b=3√,
即有椭圆C2的方程为x24+y23=1;
(Ⅱ)设l:y=k(x−1),k≠0,代入椭圆方程,可得(3+4k2)x2−8k2x+4k2−12=0,
设A(x1,y1),B(x2,y2),则x1+x2=8k23+4k2,x1x2=4k2−123+4k2,
即有y1+y2=k(x1+x2)−2k=8k33+4k2−2k=−6k3+4k2
由P为中点,可得P(4k23+4k2,−3k3+4k2),又PD的斜率为−1k,
即有PD:y−−3k3+4k2=−1k(x−4k23+4k2),令y=0,可得x=k23+4k2,
即有D(k23+4k2,0),
可得|PD|=(k23+4k2−4k23+4k2)2+(−3k3+4k2)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷=3k4+k2−−−−−√3+4k2,
又|AB|=1+k2−−−−−√⋅(x1+x2)2−4x1x2−−−−−−−−−−−−−−√=1+k2−−−−−√⋅(8k23+4k2)2−4(4k2−12)3+4k2−−−−−−−−−−−−−−−−−−−−−⎷
=12(1+k2)3+4k2,
即有|DP||AB|=14k2k2+1−−−−−√=141−11+k2−−−−−−−−√,
由k2+1>1,可得0<11+k2<1,
即有0<141−11+k2−−−−−−−−√<14,
则有|DP||AB|的取值范围为(0,14).