2018届高考数学突破五直线与圆锥曲线压轴大题5.1直线与圆及圆锥(2)

得(3k+1)x-18kx+27k-6=0, 依题意Δ=12(2-3k)>0, 得-

设P(x1,y1),Q(x2,y2), 则x1+x2=,x1x2=.

2

2222

∵F(2,0),M(x1,-y1),=(2-x1,y1),=(x2-2,y2),

由(2-x1)y2-(x2-2)y1

=(2-x1)·k(x2-3)-(x2-2)·k(x1-3) =k[5(x1+x2)-2x1x2-12] =k =0,

得,

∴M,F,Q三点共线.

(3)解 设直线PQ的方程为x=my+3.

由方程组得(m+3)y+6my+3=0, 依题意Δ=36m-12(m+3)>0,得m>. 设P(x1,y1),Q(x2,y2), 则y1+y2=-,

2

2

2

2

2

y1y2=.

∴S△FPQ=|AF|·|y1-y2| = = = =.

令t=m+3, 则S△FPQ=|y1-y2|

2

= =,

∴,t=m2+3=9,

即m=6,m=±时,S△FPQ最大,

2

∴S△FPQ最大时直线PQ的方程为x±y-3=0.