Question

已知P是抛物线y^2=8x上动点,Q(a,0)(a>0)求PQ的最小值

Answer 1

设P(x,y) 则y^2=8x 则x>=0

|PQ|^2=(x-a)^2+y^2=x^2+(8-2a)x+a^2=(x-(a-4))^2+8(a-2)

当a-4=0时,最小值在x=a-4时取得,最小值是2√2(a-2)

所以0