,,,,,,故抛物线C的方程为y'=4:t...........................................................................................(4分）(11)巾题意可知且线L,L,的斜率存在，且不为0,设直线I,:x=t(y-a)设点M(x1,y,),N(:r2,y2).'=4,联立方程可得rx悄去x.可得,.'-4ty+4at.=0,.=t(y一a),则.:11=161'.-l6at.>0....................................................................................................(5分）囚为y1+y2=4t,y1y2=4at,所以IMNI=✓口下ly1-y21=✓i了了小6(t'一at)=4✓i了了ff二元11+l.(,i焦点F到且线l,的距离d=/i'+'7所以S=上x4✓言下�x11+tal=2./,丁:W11+tal...........................................c,,-11,2(7分）jj'7il"2设且线12:-�=-t(y-a),与抛物线方程联立可得,12=161+16at>0,将tJI)-I杆换，可得SA/'l'IJ=2.ff冗tlt<.t-11...................................................................(8分）山sA,.11,=S凸,.,,。可得2✓了二盂II+tal=2✓i'了了九／（／，一II,2即《三已尸叫．两边平方井化简可1U1=-气，I-ata一I2-a所以2-a'>0,斛得0<a.<.ff........................................................................................(10分）义巾'11>0且牛>01从I<一(I或I>a,可知t'><卢2所以—气辽，即(a'一l)>0,所以a,"I,2-a所以实数o的取伯范围是(O,l)U{l,./2).........................................................................(12分）22.命题意图本题主没考众参数方程与普通方程的轧化以及=.f(I们邪变换解析(I)1111线C的许通方程为土·+/=I......................................................................(2分）4所以曲线C与y轴负半轴的交．权为(0,-I)因为八线l恒过点(I,0).义点(0,一I)在杠线l上，所以门线l的斜率为-I-0=I,......................................................,(4分）0-1一'IT..............................................................................(5所以tana=I,所以Ct=4.分）</a.<.ff........................................................................................(10分）义巾'11>
