若抛物线 $G$: $y = ax^2 + bx+ c$的图像经过$(-5, 0), left(0, displaystyle{5over2} ight), (1, 6)$ 三点. $Q(x, y)$ 是直线 $l$: $y = 2x - 3$ 上的动点, $P$ 是与 $l$ 平行的直线 $y = 2x + m$ 与抛物线 $G$ 的惟一公共点. 求 $P, Q$ 两点距离的最小值.
解答:
令 $f(x) = ax^2 + bx + c$, 由已知得 $$egin{cases}f(-5) = 0\ f(0) = displaystyle{5over2}\ f(1) = 6 end{cases}Rightarrow egin{cases}a = displaystyle{1over2}\ b = 3\ c = displaystyle{5over2} end{cases}$$ $ herefore f(x) = displaystyle{1over2}x^2 + 3x + displaystyle{5over2}$.
$P$ 是 $f(x)$ 与 $y = 2x + m$ 之切点: $$Rightarrowegin{cases} y = displaystyle{1over2}x^2 + 3x + displaystyle{5over2}\ y = 2x + mend{cases}$$ $$Rightarrow x^2 + 2x + 5 - 2m = 0$$ $$Rightarrow Delta = 4 - 4(5-2m) = 0$$ $$Rightarrow m = 2$$ $ herefore P(-1, 0)$.
$PQ$ 之最短距离即为平行线 $2x-3$ 与 $2x+2$ 之距离: $$ herefore |PQ|_ ext{min} = {|2 - (-3)| over sqrt{1 + 2^2}} = sqrt5.$$
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