每日一题_191112

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\(M(x_0,y_0)\)\((2\leqslant y_0\leqslant 4)\)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\(\left(\dfrac{p}{2},0\right)\),于是\(|EF|\)的最大值为\[ \left|4-\dfrac{p}{2} \right|+1=\dfrac{19}{4}.\]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\[ x-\left(t_0+t_1\right)y+2pt_0t_1=0.\]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\[ \left(4+2pt_0t_1\right)^2=1+\left(t_0+t_1\right)^2.\]同理,由\(MB\)与\(C_2\)相切可得\[ \left(4+2pt_0t_2\right)^2=1+\left(t_0+t_2\right)^2.\]两式作差可得\[\left[8+2pt_0\left(t_1+t_2\right)\right]\cdot\left(t_1-t_2\right)\cdot 2pt_0= \left(2t_0+t_1+t_2\right)\cdot\left(t_1-t_2\right).\]显然\(t_1-t_2\neq 0\),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\[\left[-\dfrac{2(8p-1)}{3},-\dfrac{4(8p-1)}{15} \right].\]
情形一 当\(p=\dfrac12\),\(AB\)中点的纵坐标取值范围为\(\left[-2,-\dfrac{4}{5}\right]\).
情形二 当\(p=\dfrac{31}{2}\),\(AB\)中点的纵坐标取值范围为\(\left[-82,-\dfrac{164}{5}\right]\).