2023年高考真题练习

在直角坐标系$xOy$中，点$P$到$x$轴的距离等于点$P$到点$(0,\frac{1}{2})$的距离，记动点$P$的运动轨迹为$W$
（1）求$W$的方程
（2）已知矩形$ABCD$有三个顶点在$W$上，证明：矩形的周长大于$3\sqrt{3}$

解：
\quad (1)依题意，$x^2+(y-\frac{1}{2}{)}^2=y^2$

\qquad 整理得$y=x^2+\frac{1}{4}$

\quad (2)不妨设矩形在$W$上的三个点为$A,B,C$，且$A,B$在$x$轴同侧，$AB\perp AC$

$y=x^2+\frac{1}{4}$的对称轴为$x$轴

\qquad 不妨设$A,B$都在$x$轴右边，若在左边，则将$W$关于$x$轴对称使$A,B$翻转到$x$轴右边

\qquad 设$A(p,p^2+\frac{1}{4})$，$B(q,q^2+\frac{1}{4})$，$C(t,t^2+\frac{1}{4})$

\qquad 则$AB:y=(p+q)x-pq+\frac{1}{4}$

$\because AC\perp AB$

$\therefore AC:y=-\frac{1}{p+q}x+\frac{p}{p+q}+p^2+\frac{1}{4}$

$t^2+\frac{1}{4}=-\frac{1}{p+q}t+\frac{p}{p+q}+p^2+\frac{1}{4}$，解得$t=p$（舍）或$t=-p-\frac{1}{p+q}$

$\therefore C(-p-\frac{1}{p+q},(p+\frac{1}{p+q}{)}^2+\frac{1}{4})$

$AB=\sqrt{(p-q{)}^2+(p^2-q^2{)}^2}=|p^2-q^2|\sqrt{1+\frac{1}{(p+q{)}^2}}$

$AC=\sqrt{(2p+\frac{1}{p+q}{)}^2+[\frac{2p}{p+q}+\frac{1}{(p+q{)}^2}{]}^2}=(2p+\frac{1}{p+q})\sqrt{1+\frac{1}{(p+q{)}^2}}$

\qquad 令$W=AB+AC=(|p^2-q^2|+2p+\frac{1}{p+q})\sqrt{1+\frac{1}{(p+q{)}^2}}$

\qquad 证周长大于$3\sqrt{3}$，即证$W>\frac{3\sqrt{3}}{2}$

\qquad 令$k=p+q$，则$W=(k|p-q|+2p+\frac{1}{k})\sqrt{1+\frac{1}{k^2}}$

$\because A,B$都在$x$轴右边

$\therefore k>0$

\qquad ①$0<k\le 1$时

$\because k|p-q|+2p\ge k(q-p)+2p=k(p+q)+(2-2k)p\ge k^2$

$\therefore W\ge (k^2+\frac{1}{k})\sqrt{1+\frac{1}{k^2}}$

$W^2\ge k^4+k^2+2k+\frac{2}{k}+\frac{1}{k^2}+\frac{1}{k^4}\ge 2\sqrt{k^4\cdot \frac{1}{k^4}}+2\sqrt{k^2\cdot \frac{1}{k^2}}+2\times 2\sqrt{k\cdot \frac{1}{k}}=8$

$W^2\ge 8>\frac{27}{4}$，即$W>\frac{3\sqrt{3}}{2}$

\qquad ②$k>1$时

$\because k|p-q|+2p+\frac{1}{p+q}>|p-q|+2p+\frac{1}{p+q}\ge (q-p)+2p+\frac{1}{p+q}=k+\frac{1}{k}$

$\therefore W>(k+\frac{1}{k})\sqrt{1+\frac{1}{k^2}}$，$W^2>k^2(1+\frac{1}{k^2}{)}^3$

\qquad 证明$k^2(1+\frac{1}{k^2}{)}^3\ge \frac{27}{4}$，即证$(1+\frac{1}{k^2}{)}^3\ge \frac{27}{4k^2}$

\qquad 即$1+k^{-2}\ge 3\cdot (4k^2{)}^{-\frac{1}{3}}$，$1+k^{-2}-3\cdot 4^{-\frac{1}{3}}\cdot k^{-\frac{2}{3}}\ge 0$

\qquad 令$g(x)=1+x^{-2}-3\cdot 4^{-\frac{1}{3}}\cdot x^{-\frac{2}{3}}$

\qquad 则$g^{\prime }(x)=-2x^{-3}+4^{\frac{1}{6}}{x}^{-\frac{5}{3}}$

$x=2$时$g^{\prime }(x)=0$

$\therefore g(x)$在$[\sqrt{2},+\infty )$上单调递增，在$(-\infty ,\sqrt{2})$上单调递减

$\because g(\sqrt{2})=1+\frac{1}{2}-3\cdot 4^{-\frac{1}{3}}\cdot 4^{-\frac{1}{6}}=0$

$g(x)\ge g(\sqrt{2})=0$，$x\in (1,+\infty )$

$\therefore k^2(1+\frac{1}{k^2}{)}^3\ge \frac{27}{4}$

$\because W^2>k^2(1+\frac{1}{k^2}{)}^3\ge \frac{27}{4}$

$\therefore W>\frac{3\sqrt{3}}{2}$

\qquad 综上所述，$W>\frac{3\sqrt{3}}{2}$

\qquad 得证矩形的周长大于$3\sqrt{3}$