由題意易得F1(-1,0)；F2(1,0)
∵P(x,y)
∴向量PF1=(-1-x,0-y)=(-1-x,-y)
向量PF2=(1-x,0-y)=(1-x,-y)
∴向量PF1*向量PF2=(-1-x)×(1-x)+(-y)×(-y)=x²-1+y²
由x²/4+y²/3=1得y²=3-(3/4)x²
∴向量PF1*向量PF2=(-1-x)×(1-x)+(-y)×(-y)=x²-1+y²
=x²-1+3-(3/4)x²
=(1/4)x²+2(-2≤x≤2)
∴0≤x²≤4
故向量PF1*向量PF2的范圍是[2,3]