1. 用含 $m$ 的代数式表示点 $P$、$D$ 的坐标及线段 $PD$ 的长
由题意,点 P P P 在直线 B C BC BC 下方的抛物线上,且横坐标为 m m m ,易知 0 < m < 3 0 < m < 3 0 < m < 3 .
则点 P P P 的坐标为 ( m , m 2 − 2 m − 3 ) (m, m^2 - 2m - 3) ( m , m 2 − 2 m − 3 ) .
因为 P F ⊥ x PF \perp x PF ⊥ x 轴交直线 B C BC BC 于点 D D D ,所以点 D D D 的横坐标也为 m m m ,其坐标为 ( m , m − 3 ) (m, m - 3) ( m , m − 3 ) .
故线段 P D PD P D 的长为:
P D = y D − y P = ( m − 3 ) − ( m 2 − 2 m − 3 ) = − m 2 + 3 m = m ( 3 − m ) PD = y_D - y_P = (m - 3) - (m^2 - 2m - 3) = -m^2 + 3m = m(3 - m) P D = y D − y P = ( m − 3 ) − ( m 2 − 2 m − 3 ) = − m 2 + 3 m = m ( 3 − m )
2. 分析 $\Delta PED$ 的形状并求其面积
直线 B C BC BC 的解析式为 y = x − 3 y = x - 3 y = x − 3 ,其斜率为 1 1 1 ,因而直线 B C BC BC 与 x x x 轴正方向的夹角为 45 ∘ 45^\circ 4 5 ∘ .
又因为 P D ∥ y PD \parallel y P D ∥ y 轴,所以直线 P D PD P D 与直线 B C BC BC 的夹角 ∠ P D E = 45 ∘ \angle PDE = 45^\circ ∠ P D E = 4 5 ∘ .
已知 P E ⊥ B C PE \perp BC PE ⊥ BC 于点 E E E ,所以在 Rt Δ P E D \text{Rt}\Delta PED Rt Δ PE D 中,∠ P E D = 90 ∘ \angle PED = 90^\circ ∠ PE D = 9 0 ∘ 且 ∠ P D E = 45 ∘ \angle PDE = 45^\circ ∠ P D E = 4 5 ∘ .
因此 Δ P E D \Delta PED Δ PE D 是等腰直角三角形,得:
P E = E D = 2 2 P D PE = ED = \frac{\sqrt{2}}{2} PD PE = E D = 2 2 P D
其面积为:
S Δ P E D = 1 2 P E ⋅ E D = 1 2 ( 2 2 P D ) 2 = 1 4 P D 2 = 1 4 [ m ( 3 − m ) ] 2 S_{\Delta PED} = \frac{1}{2} PE \cdot ED = \frac{1}{2} \left(\frac{\sqrt{2}}{2} PD\right)^2 = \frac{1}{4} PD^2 = \frac{1}{4} [m(3 - m)]^2 S Δ PE D = 2 1 PE ⋅ E D = 2 1 ( 2 2 P D ) 2 = 4 1 P D 2 = 4 1 [ m ( 3 − m ) ] 2
3. 求 $\Delta PDB$ 的面积及两三角形面积的比值
以 P D PD P D 为底,则点 B ( 3 , 0 ) B(3, 0) B ( 3 , 0 ) 到直线 P D PD P D (即直线 x = m x = m x = m )的距离即为高,其值为 x B − x D = 3 − m x_B - x_D = 3 - m x B − x D = 3 − m .
故 Δ P D B \Delta PDB Δ P D B 的面积为:
S Δ P D B = 1 2 P D ⋅ ( 3 − m ) = 1 2 m ( 3 − m ) ⋅ ( 3 − m ) = 1 2 m ( 3 − m ) 2 S_{\Delta PDB} = \frac{1}{2} PD \cdot (3 - m) = \frac{1}{2} m(3 - m) \cdot (3 - m) = \frac{1}{2} m(3 - m)^2 S Δ P D B = 2 1 P D ⋅ ( 3 − m ) = 2 1 m ( 3 − m ) ⋅ ( 3 − m ) = 2 1 m ( 3 − m ) 2
因此,两三角形的面积比为:
S Δ P E D S Δ P D B = 1 4 m 2 ( 3 − m ) 2 1 2 m ( 3 − m ) 2 = m 2 \frac{S_{\Delta PED}}{S_{\Delta PDB}} = \frac{\frac{1}{4} m^2 (3 - m)^2}{\frac{1}{2} m (3 - m)^2} = \frac{m}{2} S Δ P D B S Δ PE D = 2 1 m ( 3 − m ) 2 4 1 m 2 ( 3 − m ) 2 = 2 m
4. 分类讨论面积比为 $4:5$ 建立方程并求解 $m$
线段 P D PD P D 把 Δ P E B \Delta PEB Δ PEB 分成的两个三角形为 Δ P E D \Delta PED Δ PE D 和 Δ P D B \Delta PDB Δ P D B ,其面积比为 4 : 5 4:5 4 : 5 ,需分两种情况讨论:
① 若 S Δ P E D S Δ P D B = 4 5 \dfrac{S_{\Delta PED}}{S_{\Delta PDB}} = \dfrac{4}{5} S Δ P D B S Δ PE D = 5 4 ,则:
m 2 = 4 5 ⟹ m = 8 5 \frac{m}{2} = \frac{4}{5} \implies m = \frac{8}{5} 2 m = 5 4 ⟹ m = 5 8
经检验,m = 8 5 ∈ ( 0 , 3 ) m = \dfrac{8}{5} \in (0, 3) m = 5 8 ∈ ( 0 , 3 ) ,符合题意;
② 若 S Δ P D B S Δ P E D = 4 5 \dfrac{S_{\Delta PDB}}{S_{\Delta PED}} = \dfrac{4}{5} S Δ PE D S Δ P D B = 5 4 ,即 S Δ P E D S Δ P D B = 5 4 \dfrac{S_{\Delta PED}}{S_{\Delta PDB}} = \dfrac{5}{4} S Δ P D B S Δ PE D = 4 5 ,则:
m 2 = 5 4 ⟹ m = 5 2 \frac{m}{2} = \frac{5}{4} \implies m = \frac{5}{2} 2 m = 4 5 ⟹ m = 2 5
经检验,m = 5 2 ∈ ( 0 , 3 ) m = \dfrac{5}{2} \in (0, 3) m = 2 5 ∈ ( 0 , 3 ) ,符合题意.
综上所述,m m m 的值为 8 5 \dfrac{8}{5} 5 8 或 5 2 \dfrac{5}{2} 2 5 .
答案 m = 8 5 m = \dfrac{8}{5} m = 5 8 或 m = 5 2 m = \dfrac{5}{2} m = 2 5