同构处理
过点\(P\)做\(x\)轴的垂线，垂足为\(E\)，且该垂线与抛物线\(x^2=-4y\)交与点\(F\),\(|PE|^2+|EF|=1\)，记动点\(P\)的轨迹为\(C\)
\((1)\) 求出\(C\)的轨迹方程
\((2)\) 圆\(Q\)是以点\(Q(1,0)\)为圆心，\(r(0<r<1)\)为半径的圆，过点\(B(0,-1)\)作圆\(Q\)的两条切线，这两条切线分别与\(C\)相交于点\(M,N\)(异于点\(B\).)当\(r\)变化时，是否存在点\(G\)使得直线恒过定点\(G\)？
解
\((1)\) 设\(P:(x,y),F:\left(x,-\dfrac{x^2}{4}\right)\)
因\(|PE|^2+|EF|=1\)则有\(C:\dfrac{x^2}{4}+y^2=1\)
\((2)\) 设过\(BM\)直线\(y=kx-1\)
则由相切的条件得到：\(\dfrac{|-k-1|}{\sqrt{1+k^2}}=r\)
即\(k^2(1-r^2)+2k+1-r^2=0\)
则\(k_{MB}\cdot k_{NB}=1\)
设\(BN:y=k_{1}x-1,BM:y=k_{2}-x\)
即\(k_{1}-(1+y)=0,k_{2}-(1+y)=0\)
两直线左右两边相乘有：\(k_1k_2x^2-(1+y)x(k_1+k_2)+(1+y)^2=0\)
联立\(C:\dfrac{x^2}{4}+y^2=1\)有
\(4k_1k_2(1-y)(1+y)-(1+y)x(k_1+k_2)+(1+y)^2=0\)
因\(k_1k_2=1\)，从而\(MN\)直线为
\(4(1-y)-x(k_1+k_2)+1+y=0\)
过定点\(\left(0,\dfrac{5}{3}\right)\)