若抛物线 $G$: $y = ax^2 + bx+ c$的图像经过$(-5, 0), \left(0, \displaystyle{5\over2}\right), (1, 6)$ 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 $$\begin{cases}f(-5) = 0\\ f(0) = \displaystyle{5\over2}\\ f(1) = 6 \end{cases}\Rightarrow \begin{cases}a = \displaystyle{1\over2}\\ b = 3\\ c = \displaystyle{5\over2} \end{cases}$$ $\therefore f(x) = \displaystyle{1\over2}x^2 + 3x + \displaystyle{5\over2}$.
$P$ 是 $f(x)$ 与 $y = 2x + m$ 之切点: $$\Rightarrow\begin{cases} y = \displaystyle{1\over2}x^2 + 3x + \displaystyle{5\over2}\\ y = 2x + m\end{cases}$$ $$\Rightarrow x^2 + 2x + 5 - 2m = 0$$ $$\Rightarrow \Delta = 4 - 4(5-2m) = 0$$ $$\Rightarrow m = 2$$ $\therefore P(-1, 0)$.
$PQ$ 之最短距离即为平行线 $2x-3$ 与 $2x+2$ 之距离: $$\therefore |PQ|_\text{min} = {|2 - (-3)| \over \sqrt{1 + 2^2}} = \sqrt5.$$
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