每日一题_191112

Posted 2020-11-30 math521

已知(F)为抛物线(C_1:y^2=2px)((p>0))的焦点,(E)为圆(C_2:(x-4)^2+y^2=1)上任意一点,且(|EF|)的最大值为(dfrac{19}{4}).
((1)) 求抛物线(C_1)的方程;
((2)) 若(M(x_0,y_0))((2leqslant y_0leqslant 4))在抛物线(C_1)上,过(M)作圆(C_2)的两条切线,交抛物线(C_1)于(A,B),求(AB)中点的纵坐标的取值范围.
解析:
((1)) 由题易知(F)的坐标为(left(dfrac{p}{2},0 ight)),于是(|EF|)的最大值为[ left|4-dfrac{p}{2} ight|+1=dfrac{19}{4}.]解得(p=dfrac{1}{2})或(p=dfrac{31}{2}).所以所求抛物线方程为(y^2=x)或(y^2=31x).
((2)) 由题,设[M(2pt_0^2,2pt_0),A(2pt_1^2,2pt_1),B(2pt_2^2,2pt_2).]易得直线(MA)的一般方程为[ x-left(t_0+t_1 ight)y+2pt_0t_1=0.]由于直线(MA)与圆(C_2)相切,所以圆心(C_2)到直线(MA)的距离为(1),即有[ left(4+2pt_0t_1 ight)^2=1+left(t_0+t_1 ight)^2.]同理,由(MB)与(C_2)相切可得[ left(4+2pt_0t_2 ight)^2=1+left(t_0+t_2 ight)^2.]两式作差可得[left[8+2pt_0left(t_1+t_2 ight) ight]cdotleft(t_1-t_2 ight)cdot 2pt_0= left(2t_0+t_1+t_2 ight)cdotleft(t_1-t_2 ight).]显然(t_1-t_2 eq 0),若记(AB)中点横坐标为(m),则[ m=dfrac{2pt_1+2pt_2}{2}=dfrac{16p^2t_0-2pt_0}{1-4p^2t_0^2}=dfrac{(8p-1)y_0}{1-y_0^2}.]无论(p=dfrac{1}{2})还是(dfrac{31}{2}),(m)都是关于(y_0)的单调递增函数,因此(m)的取值范围为[left[-dfrac{2(8p-1)}{3},-dfrac{4(8p-1)}{15} ight].]
情形一 当(p=dfrac12),(AB)中点的纵坐标取值范围为(left[-2,-dfrac{4}{5} ight]).
情形二 当(p=dfrac{31}{2}),(AB)中点的纵坐标取值范围为(left[-82,-dfrac{164}{5} ight]).