《2025年实验班提优训练九年级数学上册人教版》

15. 分类讨论思想(2023·南充中考)如图(1),抛物线$y= ax^{2}+bx+3(a≠0)$与x轴交于$A(-1,0),$$B(3,0)$两点,与y轴交于点C.
(1)求抛物线的解析式;
(2)点P在抛物线上,点Q在x轴上,以B,C,P,Q为顶点的四边形为平行四边形,求点P的坐标;
(3)如图(2),抛物线顶点为D,对称轴与x轴交于点E,过点$K(1,3)$的直线(直线KD除外)与抛物线交于G,H两点,直线DG,DH分别交x轴于点M,N.试探究$EM\cdot EN$是否为定值,若是,求出该定值;若不是,说明理由.

(1)抛物线的解析式为$y=-x²+2x+3$.
(2)点P的坐标为$(2,3)$或$(1+\sqrt{7},-3)$或$(1-\sqrt{7},-3)$.
(3)EM·EN是定值,该定值为16.

答案：
(1)由题意,得抛物线的解析式为y = a(x + 1)(x - 3) = a(x² - 2x - 3) = ax² - 2ax - 3a,
∴ - 3a = 3,
∴a = - 1,
∴抛物线的解析式为y = - x² + 2x + 3.
(2)设点P的坐标为(t,-t² + 2t + 3),点Q的坐标为(x,0).当BC或BP为对角线时,由中点坐标公式,得3 = - t² + 2t + 3,解得t = 0(舍去)或2,
∴点P的坐标为(2,3).当BQ为对角线时,同理可得0 = - t² + 2t + 3 + 3,解得t = 1 ± $\sqrt{7}$,
∴点P的坐标为(1 + $\sqrt{7}$,-3)或(1 - $\sqrt{7}$,-3).综上所述,点P的坐标为(2,3)或(1 + $\sqrt{7}$,-3)或(1 - $\sqrt{7}$,-3).
(3)EM·EN是定值.理由如下:由直线GH过点K(1,3),可设直线GH的表达式为y = k(x - 1) + 3,设点G,H的坐标分别为(m,-m² + 2m + 3)和(n,-n² + 2n + 3),联立y = k(x - 1) + 3和y = - x² + 2x + 3并整理,得x² + (k - 2)x - k = 0,
∴m + n = 2 - k,mn = - k.
∵y = - x² + 2x + 3 = -(x - 1)² + 4,
∴D(1,4).由点G,D的坐标易得直线GD的表达式为y = -(m - 1)(x - 1) + 4,令y = 0,则x = 1 + $\frac{4}{m - 1}$,即点M(1 + $\frac{4}{m - 1}$,0),则EM = 1 - 1 - $\frac{4}{m - 1}$ = -$\frac{4}{m - 1}$,同理可得EN = $\frac{4}{n - 1}$,则EM·EN = -$\frac{4}{m - 1}$×$\frac{4}{n - 1}$ = -$\frac{16}{(m - 1)(n - 1)}$ = $\frac{-16}{mn - (m + n) + 1}$ = $\frac{-16}{-k + k - 2 + 1}$ = 16.
思路引导　
(1)由待定系数法即可求解;
(2)当BC或BP为对角线时,由中点坐标公式列出等式,即可求解;当BQ为对角线时,同理可解;
(3)求出直线GD的表达式为y = -(m - 1)(x - 1) + 4,得到点M(1 + $\frac{4}{m - 1}$,0),同理可得EN = $\frac{4}{n - 1}$,即可求解.

16. 中考新考法 定值问题探究 (2024·宿迁中考)如图(1),已知抛物线$y_{1}= x^{2}+bx+c$与x轴交于两点$O(0,0),A(2,0)$,将抛物线$y_{1}$向右平移两个单位长度,得到抛物线$y_{2}$.点P是抛物线$y_{1}$在第四象限内一点,连接PA并延长,交抛物线$y_{2}$于点Q.
(1)求抛物线$y_{2}$的表达式;$y_{2}=x^{2}-6x+8$
(2)设点P的横坐标为$x_{P}$,点Q的横坐标为$x_{Q}$,求$x_{Q}-x_{P}$的值;4
(3)如图(2),若抛物线$y_{3}= x^{2}-8x+t与抛物线y_{1}= x^{2}+bx+c$交于点C,过点C作直线MN,分别交抛物线$y_{1}和y_{3}$于点M,N(M,N均不与点C重合),设点M的横坐标为m,点N的横坐标为n,试判断$|m-n|$是否为定值.若是,直接写出这个定值;若不是,请说明理由.是，6

答案：
(1)由题意,得y₁ = x(x - 2) = x² - 2x.将抛物线y₁向右平移两个单位长度得到抛物线y₂,则y₂过点(2,0),(4,0),
∴y₂ = (x - 2)(x - 4) = x² - 6x + 8.
(2)设点P(p,p² - 2p),直线PA的表达式为y = k(x - 2),将点P的坐标代入,得p² - 2p = k(p - 2),解得k = p,
∴直线AP的表达式为y = p(x - 2),联立上式和抛物线y₂的表达式,得x² - 6x + 8 = p(x - 2),解得xQ = 4 + p或xQ = 2(舍去),
∴xQ - xP = 4 + p - p = 4.
(3)由
(1)知y₁ = x(x - 2) = x² - 2x,
∴M(m,m² - 2m),联立y₁,y₃得x² - 2x = x² - 8x + t,解得x = $\frac{1}{6}t$,
∴点C($\frac{1}{6}t$,$\frac{1}{36}t² - \frac{1}{3}t$).由点C,M的坐标,得直线CM的表达式为y = (m + $\frac{1}{6}t$ - 2)(x - m) + m² - 2m,联立上式和y₃的表达式,得x² - 8x + t = (m + $\frac{1}{6}t$ - 2)(x - m) + m² - 2m,整理,得x² - (6 + m + $\frac{1}{6}t$)x + (1 + $\frac{1}{6}m$)t = 0,则xC + xN = 6 + m + $\frac{1}{6}t$,即$\frac{1}{6}t$ + n = 6 + m + $\frac{1}{6}t$,
∴n - m = 6,
∴|m - n| = 6为定值.