在△ABC中,AC=2√3,B是椭圆x^2/5+y^2/4=1在x轴上方的顶点,l是双曲线x^2-y^2=-2位于x轴下

B(0,2), 双曲线:y?/2-x?/2=1, l:y=-2/2=-11)设A(t-√3,-1),C(t+√3,-1), AC中垂线为x=tAB斜率为: (2+1)/(-t+√3)=3/(√3-t)∴AB中垂线斜率为 -1/[3/(√3-t)]=(t-√3)/3AB中垂线过AB中点((t-√3)/2,1/2)∴AB中垂线: y=(t-√3)/3×[x-(t-√3)/2]+1/2P为AB中垂线和AC中垂线x=t的交点, 令x=t. 则y=(t-√3)/3[t-(t-√3)/2]-1/2=(t-√3)(t+√3)/6-1/2=(t?-3)/6-1/2=t?/6∴P(t,t?/6)即P的轨迹E为 y=x?/6, 即x?=6y2)焦点F(0,3/2), 显然MN,RQ的斜率都存在设MN斜率为k, RQ斜率为-1/k, 则MN:y=kx+3/2, RQ:y=-x/k+3/2将MN代入抛物线得 x?=6y=6(kx+3/2)x?-6kx-9=0, x1+x2=6k, x1x2=-9∴|x1-x2|=√[(x1+x2)?-4x1x2]=√(36k?+36)=6√(1+k?)∴MN=√(1+k?)×|x1-x2|=√(1+k?)×6√(1+k?)=6(1+k?), 同理RQ=6(1+1/k?)S=(1/2)MN×RQ=18(1+k?)(1+1/k?)=18(1+k?+1/k?+1)=18(2+k?+1/k?)>=18(2+2√(k?×1/k?))=18(2+2)=72取等k?=1/k?, k=1或-1, 即MN,PQ的斜率分别为1或-1综上,MRNQ的面积最小值为72