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在直角坐标系 x O y xOy xOy中，点 P P P到 x x x轴的距离等于点 P P P到点 ( 0 , 1 2 ) (0,\dfrac 12) (
0,21)的距离，记动点 P P P的运动轨迹为 W W W
（1）求 W W W的方程
（2）已知矩形 A B C D ABCD ABCD有三个顶点在 W W W上，证明：矩形的周长大于 3 3 3\sqrt 3 33
解：
\quad (1)依题意， x 2 + ( y − 1 2 ) 2 = y 2 x^2+(y-\dfrac 12)^2=y^2 x2+(y−21)2=y2
\qquad 整理得 y = x 2 + 1 4 y=x^2+\dfrac 14 y=x2+41
\quad (2)不妨设矩形在 W W W上的三个点为 A , B , C A,B,C A,B,C，且 A , B A,B A,B在 x x x轴同侧，
A B ⊥ A C AB\bot ACAB⊥AC
y = x 2 + 1 4 \qquad y=x^2+\dfrac 14 y=x2+41的对称轴为 x x x轴
\qquad 不妨设 A , B A,B A,B都在 x x x轴右边，若在左边，则将 W W W关于 x x x轴对称使 A , B A,B A,B翻转到
x xx轴右边
\qquad 设 A ( p , p 2 + 1 4 ) A(p,p^2+\dfrac 14) A(p,p2+41)， B ( q , q 2 + 1
4 ) B(q,q^2+\dfrac14)B(q,q2+41)， C ( t , t 2 + 1 4 ) C(t,t^2+\dfrac14) C(t,t2+4
1)
\qquad 则 A B : y = ( p + q ) x − p q + 1 4 AB:y=(p+q)x-pq+\dfrac 14 AB:y=(p+q)
x−pq+41
∵ A C ⊥ A B \qquad \because AC\bot AB ∵AC⊥AB
∴ A C : y = − 1 p + q x + p p + q + p 2 + 1 4 \qquad \therefore
AC:y=-\dfrac{1}{p+q}x+\dfrac{p}{p+q}+p^2+\dfrac 14∴AC:y=−p+q1x+p+qp+p2+41
t 2 + 1 4 = − 1 p + q t + p p + q + p 2 + 1 4 \qquad t^2+\dfrac
14=-\dfrac{1}{p+q}t+\dfrac{p}{p+q}+p^2+\dfrac 14t2+41=−p+q1t+p+qp+p2+41，解得
t = p t=pt=p（舍）或 t = − p − 1 p + q t=-p-\dfrac{1}{p+q} t=−p−p+q1
∴ C ( − p − 1 p + q , ( p + 1 p + q ) 2 + 1 4 ) \qquad \therefore
C(-p-\dfrac{1}{p+q},(p+\dfrac{1}{p+q})^2+\dfrac 14)∴C(−p−p+q1,(p+p+q1)2+41)
A B = ( p − q ) 2 + ( p 2 − q 2 ) 2 = ∣ p 2 − q 2 ∣ 1 + 1 ( p + q ) 2 \qquad
AB=\sqrt{(p-q)^2+(p^2-q^2)^2}=|p^2-q^2|\sqrt{1+\dfrac{1}{(p+q)^2}}AB=(p−q)2+(p2−
q2)2 =∣p2−q2∣1+(p+q)21
A C = ( 2 p + 1 p + q ) 2 + [ 2 p p + q + 1 ( p + q ) 2 ] 2 = ( 2 p + 1 p + q
) 1 + 1 ( p + q ) 2 \qquad
AC=\sqrt{(2p+\dfrac{1}{p+q})^2+[\dfrac{2p}{p+q}+\dfrac{1}{(p+q)^2}]^2}=(2p+\dfrac{1}{p+q})\sqrt{1+\dfrac{1}{(p+q)^2}}
AC=(2p+p+q1)2+[p+q2p+(p+q)21]2 =(2p+p+q1)1+(p+q)21
\qquad 令 W = A B + A C = ( ∣ p 2 − q 2 ∣ + 2 p + 1 p + q ) 1 + 1 ( p + q ) 2
W=AB+AC=(|p^2-q^2|+2p+\dfrac{1}{p+q})\sqrt{1+\dfrac{1}{(p+q)^2}}W=AB+AC=(∣p2−q2∣
+2p+p+q1)1+(p+q)21
\qquad 证周长大于 3 3 3\sqrt 3 33 ，即证 W > 3 3 2 W>\dfrac{3\sqrt 3}{2} W>233
\qquad 令 k = p + q k=p+q k=p+q，则 W = ( k ∣ p − q ∣ + 2 p + 1 k ) 1 + 1 k 2
W=(k|p-q|+2p+\dfrac 1k)\sqrt{1+\dfrac{1}{k^2}}W=(k∣p−q∣+2p+k1)1+k21
∵ A , B \qquad\because A,B ∵A,B都在 x x x轴右边
∴ k > 0 \qquad\therefore k>0 ∴k>0
\qquad ① 0 < k ≤ 1 0<k\leq 1 0<k≤1时
∵ k ∣ p − q ∣ + 2 p ≥ k ( q − p ) + 2 p = k ( p + q ) + ( 2 − 2 k ) p ≥ k 2
\qquad\because k|p-q|+2p\geq k(q-p)+2p=k(p+q)+(2-2k)p\geq k^2∵k∣p−q∣+2p≥k(q−p)+2
p=k(p+q)+(2−2k)p≥k2
∴ W ≥ ( k 2 + 1 k ) 1 + 1 k 2 \qquad\therefore W\geq(k^2+\dfrac
1k)\sqrt{1+\dfrac{1}{k^2}}∴W≥(k2+k1)1+k21
W 2 ≥ k 4 + k 2 + 2 k + 2 k + 1 k 2 + 1 k 4 ≥ 2 k 4 ⋅ 1 k 4 + 2 k 2 ⋅ 1 k 2 +
2 × 2 k ⋅ 1 k = 8 \qquad W^2\geq k^4+k^2+2k+\dfrac
2k+\dfrac{1}{k^2}+\dfrac{1}{k^4}\geq
2\sqrt{k^4\cdot\dfrac{1}{k^4}}+2\sqrt{k^2\cdot\dfrac{1}{k^2}}+2\times
2\sqrt{k\cdot\dfrac 1k}=8W2≥k4+k2+2k+k2+k21+k41≥2k4⋅k41 +2k2⋅k21 +2×2k⋅k1
=8
W 2 ≥ 8 > 27 4 \qquad W^2\geq 8>\dfrac{27}{4} W2≥8>427，即 W > 3 3 2
W>\dfrac{3\sqrt 3}{2}W>233
\qquad ② k > 1 k>1 k>1时
∵ k ∣ p − q ∣ + 2 p + 1 p + q > ∣ p − q ∣ + 2 p + 1 p + q ≥ ( q − p ) + 2 p +
1 p + q = k + 1 k \qquad\because
k|p-q|+2p+\dfrac{1}{p+q}>|p-q|+2p+\dfrac{1}{p+q}\geq(q-p)+2p+\dfrac{1}{p+q}=k+\dfrac
1k∵k∣p−q∣+2p+p+q1>∣p−q∣+2p+p+q1≥(q−p)+2p+p+q1=k+k1
∴ W > ( k + 1 k ) 1 + 1 k 2 \qquad\therefore W>(k+\dfrac
1k)\sqrt{1+\dfrac{1}{k^2}}∴W>(k+k1)1+k21 ， W 2 > k 2 ( 1 + 1 k 2 ) 3
W^2>k^2(1+\dfrac{1}{k^2})^3W2>k2(1+k21)3
\qquad 证明 k 2 ( 1 + 1 k 2 ) 3 ≥ 27 4 k^2(1+\dfrac{1}{k^2})^3\geq\dfrac{27}{4}
k2(1+k21)3≥427，即证 ( 1 + 1 k 2 ) 3 ≥ 27 4 k 2
(1+\dfrac{1}{k^2})^3\geq\dfrac{27}{4k^2}(1+k21)3≥4k227
\qquad 即 1 + k − 2 ≥ 3 ⋅ ( 4 k 2 ) − 1 3 1+k^{-2}\geq3\cdot(4k^2)^{-\frac 13}
1+k−2≥3⋅(4k2)−31， 1 + k − 2 − 3 ⋅ 4 − 1 3 ⋅ k − 2 3 ≥ 0 1+k^{-2}-3\cdot
4^{-\frac 13}\cdot k^{-\frac 23}\geq 01+k−2−3⋅4−31⋅k−32≥0
\qquad 令 g ( x ) = 1 + x − 2 − 3 ⋅ 4 − 1 3 ⋅ x − 2 3 g(x)=1+x^{-2}-3\cdot
4^{-\frac 13}\cdot x^{-\frac 23}g(x)=1+x−2−3⋅4−31⋅x−32
\qquad 则 g ′ ( x ) = − 2 x − 3 + 4 1 6 x − 5 3 g'(x)=-2x^{-3}+4^{\frac
16}x^{-\frac 53}g′(x)=−2x−3+461x−35
x = 2 \qquad x=2 x=2时 g ′ ( x ) = 0 g'(x)=0 g′(x)=0
∴ g ( x ) \qquad\therefore g(x) ∴g(x)在 [ 2 , + ∞ ) [\sqrt 2,+\infty) [2 ,+∞)
上单调递增，在 ( − ∞ , 2 ) (-\infty,\sqrt 2) (−∞,2 )上单调递减
∵ g ( 2 ) = 1 + 1 2 − 3 ⋅ 4 − 1 3 ⋅ 4 − 1 6 = 0 \qquad\because g(\sqrt
2)=1+\dfrac 12-3\cdot4^{-\frac 13}\cdot 4^{-\frac 16}=0∵g(2 )=1+21−3⋅4−31⋅4−6
1=0
g ( x ) ≥ g ( 2 ) = 0 \qquad g(x)\geq g(\sqrt 2)=0 g(x)≥g(2 )=0， x ∈ ( 1 , +
∞ ) x\in(1,+\infty)x∈(1,+∞)
∴ k 2 ( 1 + 1 k 2 ) 3 ≥ 27 4 \qquad\therefore
k^2(1+\dfrac{1}{k^2})^3\geq\dfrac{27}{4}∴k2(1+k21)3≥427
∵ W 2 > k 2 ( 1 + 1 k 2 ) 3 ≥ 27 4 \qquad\because
W^2>k^2(1+\dfrac{1}{k^2})^3\geq\dfrac{27}{4}∵W2>k2(1+k21)3≥427
∴ W > 3 3 2 \qquad\therefore W>\dfrac{3\sqrt 3}{2} ∴W>233
\qquad 综上所述， W > 3 3 2 W>\dfrac{3\sqrt 3}{2} W>233
\qquad 得证矩形的周长大于 3 3 3\sqrt 3 33