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25. (12分)如图1,过$\triangle ABC$的三个顶点分别作出与水平线垂直的三条直线,外侧两条直线之间的距离叫做$\triangle ABC$的“水平宽$(a)$”,中间的这条直线在$\triangle ABC内部的线段的长度叫做\triangle ABC$的“铅垂高$(h)$”.我们可得出一种计算三角形面积的新方法：$S_{\triangle ABC}= \frac{1}{2}ah$,即三角形的面积等于水平宽与铅垂高乘积的一半.
如图2,抛物线顶点为点$C(1,4)$,交$x轴于点A(3,0)$,交$y轴于点B$.
解答下列问题：
(1)求抛物线和直线$AB$的解析式；
(2)求$\triangle CAB的铅垂高CD及S_{\triangle CAB}$；
(3)在抛物线上是否存在一点$P$,使$S_{\triangle PAB}= \frac{9}{8}S_{\triangle CAB}$？若存在,求出点$P$的坐标；若不存在,请说明理由.
如图2,抛物线顶点为点$C(1,4)$,交$x轴于点A(3,0)$,交$y轴于点B$.
解答下列问题：
(1)求抛物线和直线$AB$的解析式；
(2)求$\triangle CAB的铅垂高CD及S_{\triangle CAB}$；
(3)在抛物线上是否存在一点$P$,使$S_{\triangle PAB}= \frac{9}{8}S_{\triangle CAB}$？若存在,求出点$P$的坐标；若不存在,请说明理由.
答案：
(1)设抛物线的解析式为y₁ = a(x - 1)² + 4,将A(3,0)代入y₁ = a(x - 1)² + 4,得0 = a(3 - 1)² + 4,解得a = - 1,所以抛物线的解析式为y₁ = - (x - 1)² + 4 = - x² + 2x + 3.设直线AB的解析式为y₂ = kx + b.因为抛物线y₁ = - x² + 2x + 3与y轴交于点B,所以B点的坐标为(0,3).将A(3,0),B(0,3)代入y₂ = kx + b,得{ 0 = 3k + b,3 = b },解得{ k = - 1,b = 3 },所以直线AB的解析式为y₂ = - x + 3.
(2)因为C点坐标为(1,4),抛物线y₁ = - x² + 2x + 3,直线y₂ = - x + 3,所以当x = 1时,y₁ = 4,y₂ = 2,即C(1,4),D(1,2),所以CD = yC - yD = 4 - 2 = 2.因为A(3,0),B(0,3),所以S△CAB = 1/2CD·(xA - xB) = 1/2×2×3 = 3.
(3)存在一点P,使S△PAB = 9/8S△CAB.由
(2),知S△CAB = 3,所以S△PAB = 9/8S△CAB = 9/8×3 = 27/8.设P(m, - m² + 2m + 3).①当点P在第一象限时,过点P作PF⊥x轴交AB于点F，连接BP，AP，如图所示:则F(m, - m + 3),所以S△PAB = 27/8 = 1/2PF·(xA - xB) = 1/2×( - m² + 2m + 3 + m - 3)×3 = - 3/2m² + 9/2m,整理,得4m² - 12m + 9 = 0,即(2m - 3)² = 0,解得m₁ = m₂ = 3/2,所以P(3/2,15/4);②当点P在第二象限时,过点P作PF⊥x轴于点F，连接BP,AP，如图所示:则F(m,0)，所以S△PAB = 27/8 = S△AOB + S梯形PFOB - S△PAF = 1/2OA·OB + 1/2(PF + OB)·OF - 1/2PF·FA = 1/2×3×3 + 1/2( - m² + 2m + 3 + 3)×(0 - m) - 1/2( - m² + 2m + 3)(3 - m) = 3/2m² - 9/2m,即4m² - 12m - 9 = 0,解得m₁ = 3 + 3√2/2,m₂ = 3 - 3√2/2.由于第二象限中，点的横坐标为负值，则取m = 3 - 3√2/2,所以P(3 - 3√2/2, - 3 + 6√2/4);③当点P在第三或第四象限时，情况相同，不妨令点P在第三象限.过点A作GF⊥x轴，过点B作BG⊥GF于点G,过点P作PF⊥GF于点F，连接BP,AP，如图所示:则F(3, - m² + 2m + 3),G(3,3),所以S△PAB = 27/8 = S梯形PFGB - S△PAF - S△AGB = 1/2(PF + GB)·GF - 1/2PF·FA - 1/2GA·GB = 1/2[(3 - m) + 3]×[3 - ( - m² + 2m + 3)] - 1/2×(3 - m)[0 - ( - m² + 2m + 3)] - 1/2×3×3 = 3/2m² - 9/2m,即4m² - 12m - 9 = 0,解得m₁ = 3 + 3√2/2,m₂ = 3 - 3√2/2.当m = 3 - 3√2/2时, - m² + 2m + 3 = - 3 + 6√2/4＞0,所以点P不可能在第三象限;当m = 3 + 3√2/2时， - m² + 2m + 3 = - 3 - 6√2/4＜0,所以点P在第四象限.所以P(3 + 3√2/2, - 3 - 6√2/4).综上所述，存在一点P,使S△PAB = 9/8S△CAB,点P的坐标为(3/2,15/4)或(3 - 3√2/2, - 3 + 6√2/4)或(3 + 3√2/2, - 3 - 6√2/4).
(1)设抛物线的解析式为y₁ = a(x - 1)² + 4,将A(3,0)代入y₁ = a(x - 1)² + 4,得0 = a(3 - 1)² + 4,解得a = - 1,所以抛物线的解析式为y₁ = - (x - 1)² + 4 = - x² + 2x + 3.设直线AB的解析式为y₂ = kx + b.因为抛物线y₁ = - x² + 2x + 3与y轴交于点B,所以B点的坐标为(0,3).将A(3,0),B(0,3)代入y₂ = kx + b,得{ 0 = 3k + b,3 = b },解得{ k = - 1,b = 3 },所以直线AB的解析式为y₂ = - x + 3.
(2)因为C点坐标为(1,4),抛物线y₁ = - x² + 2x + 3,直线y₂ = - x + 3,所以当x = 1时,y₁ = 4,y₂ = 2,即C(1,4),D(1,2),所以CD = yC - yD = 4 - 2 = 2.因为A(3,0),B(0,3),所以S△CAB = 1/2CD·(xA - xB) = 1/2×2×3 = 3.
(3)存在一点P,使S△PAB = 9/8S△CAB.由
(2),知S△CAB = 3,所以S△PAB = 9/8S△CAB = 9/8×3 = 27/8.设P(m, - m² + 2m + 3).①当点P在第一象限时,过点P作PF⊥x轴交AB于点F，连接BP，AP，如图所示:则F(m, - m + 3),所以S△PAB = 27/8 = 1/2PF·(xA - xB) = 1/2×( - m² + 2m + 3 + m - 3)×3 = - 3/2m² + 9/2m,整理,得4m² - 12m + 9 = 0,即(2m - 3)² = 0,解得m₁ = m₂ = 3/2,所以P(3/2,15/4);②当点P在第二象限时,过点P作PF⊥x轴于点F，连接BP,AP，如图所示:则F(m,0)，所以S△PAB = 27/8 = S△AOB + S梯形PFOB - S△PAF = 1/2OA·OB + 1/2(PF + OB)·OF - 1/2PF·FA = 1/2×3×3 + 1/2( - m² + 2m + 3 + 3)×(0 - m) - 1/2( - m² + 2m + 3)(3 - m) = 3/2m² - 9/2m,即4m² - 12m - 9 = 0,解得m₁ = 3 + 3√2/2,m₂ = 3 - 3√2/2.由于第二象限中，点的横坐标为负值，则取m = 3 - 3√2/2,所以P(3 - 3√2/2, - 3 + 6√2/4);③当点P在第三或第四象限时，情况相同，不妨令点P在第三象限.过点A作GF⊥x轴，过点B作BG⊥GF于点G,过点P作PF⊥GF于点F，连接BP,AP，如图所示:则F(3, - m² + 2m + 3),G(3,3),所以S△PAB = 27/8 = S梯形PFGB - S△PAF - S△AGB = 1/2(PF + GB)·GF - 1/2PF·FA - 1/2GA·GB = 1/2[(3 - m) + 3]×[3 - ( - m² + 2m + 3)] - 1/2×(3 - m)[0 - ( - m² + 2m + 3)] - 1/2×3×3 = 3/2m² - 9/2m,即4m² - 12m - 9 = 0,解得m₁ = 3 + 3√2/2,m₂ = 3 - 3√2/2.当m = 3 - 3√2/2时, - m² + 2m + 3 = - 3 + 6√2/4＞0,所以点P不可能在第三象限;当m = 3 + 3√2/2时， - m² + 2m + 3 = - 3 - 6√2/4＜0,所以点P在第四象限.所以P(3 + 3√2/2, - 3 - 6√2/4).综上所述，存在一点P,使S△PAB = 9/8S△CAB,点P的坐标为(3/2,15/4)或(3 - 3√2/2, - 3 + 6√2/4)或(3 + 3√2/2, - 3 - 6√2/4).
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