设P(t,t^2+1),
则|PA|^2=(t-0)^2+(t^2+1-4)^2
=2t^2-6t+9=2(t-3/2)^2+9/2≥9/2,
所以|PA|的最小值为(3√2)/2.
已知A(0,4),P是抛物线y=x^2+1上任意一点,求|PA|的最小值.
已知A(0,4),P是抛物线y=x^2+1上任意一点,求|PA|的最小值.
数学人气:521 ℃时间:2020-04-04 02:28:05
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