《2026年金考卷中考试题汇编45套数学山东专版》

21. (本小题满分10分)
已知反比例函数$y = \frac{k}{x}(x ＞ 0)$的图象与正比例函数$y = x$的图象交于点$A(2,a)$,点$P$在线段$OA$的延长线上.
(1)求反比例函数的表达式;
(2)如图(1),过点$P$作$y$轴的平行线$l$,$l$与$y = \frac{k}{x}(x ＞ 0)$的图象交于点$B$,与$x$轴交于点$C$,当$PB = \frac{3}{4}OC$时,求点$B$的坐标;
(3)在(2)的条件下,如图(2),连接$AB$并延长,与$x$轴交于点$D$,点$Q$为$x$轴上一点,且满足$∠AQO = ∠ADO + ∠OPC$,求点$Q$的坐标.

答案： 21
(1)$\because$正比例函数$y = x$的图象过点$A(2,a)$,$\therefore a = 2$,$\therefore$点$A(2,2)$.
$\because$反比例函数$y = \frac{k}{x}(x ＞ 0)$的图象过点$A(2,2)$,$\therefore 2 = \frac{k}{2}$,$\therefore k = 4$,$\therefore$反比例函数的表达式为$y = \frac{4}{x}$.(3分)
(2)$\because$点$P$在线段$OA$的延长线上,$\therefore$设点$P(m,m)$,其中$m ＞ 2$.$\because l // y$轴,$l$与$y = \frac{k}{x}(x ＞ 0)$的图象交于点$B$,与$x$轴交于点$C$,$\therefore B(m,\frac{4}{m})$,$C(m,0)$,$\therefore PB = m - \frac{4}{m}$,$OC = m$.$\because PB = \frac{3}{4}OC$,$\therefore m - \frac{4}{m} = \frac{3}{4}m$,(4分)
解得$m = 4$或$m = -4$(已检验).
$\because m ＞ 2$,$\therefore m = 4$,$\therefore$点$B$的坐标为$(4,1)$.(6分)
(3)方法一:由
(2)得$CO = 4$,$CP = 4$,$\therefore CO = CP$.$\because \angle PCO = 90^{\circ}$,$\therefore \angle COP = \angle OPC = 45^{\circ}$,$\therefore \angle AQO = \angle ADO + \angle OPC = \angle ADO + \angle DAQ$,(7分)
$\therefore \angle DAQ = \angle DOA$.
又$\because \angle ADQ = \angle ODA$,
$\therefore \triangle ADQ \sim \triangle ODA$(点拨:共角共边反“A”字型相似),
$\therefore \frac{AD}{OD} = \frac{QD}{AD}$.
$\because A(2,2),B(4,1)$,
$\therefore$由待定系数法可得,直线$AB$的表达式为$y = - \frac{1}{2}x + 3$.
当$y = 0$时,$x = 6$,$\therefore D(6,0)$,$\therefore AD = 2\sqrt{5}$,$OD = 6$,
$\therefore \frac{2\sqrt{5}}{6} = \frac{QD}{2\sqrt{5}}$,解得$QD = \frac{10}{3}$,(9分)
$\therefore OQ = OD - QD = 6 - \frac{10}{3} = \frac{8}{3}$,
$\therefore$点$Q$的坐标为$(\frac{8}{3},0)$.(10分)
方法二:由
(2)得$CO = 4$,$CP = 4$,$\therefore CO = CP$.
$\because \angle PCO = 90^{\circ}$,$\therefore \angle POC = \angle CPO = 45^{\circ}$.
$\because \angle AQO = \angle ADO + \angle OPC$,$\angle AQO = \angle ADO + \angle DAQ$,
$\therefore \angle DAQ = \angle DOA$.
又$\because \angle ADQ = \angle ODA$,
$\therefore \triangle ADQ \sim \triangle ODA$(点拨:共角共边反“A”字型相似),
$\therefore \frac{AD}{OD} = \frac{QD}{AD}$.
$\because A(2,2),B(4,1)$,
$\therefore$由待定系数法可得,直线$AB$的表达式为$y = - \frac{1}{2}x + 3$.
当$y = 0$时,$x = 6$,$\therefore D(6,0)$,$\therefore AD = 2\sqrt{5}$,$OD = 6$,
$\therefore \frac{2\sqrt{5}}{6} = \frac{QD}{2\sqrt{5}}$,解得$QD = \frac{10}{3}$,
$\therefore OQ = OD - QD = 6 - \frac{10}{3} = \frac{8}{3}$,
$\therefore$点$Q$的坐标为$(\frac{8}{3},0)$.(10分)

22. (本小题满分11分)
在平面直角坐标系$xOy$中,抛物线$C_{1}:y = x^{2} + bx + c$经过点$A(1,0)$,$B(3,0)$,与$y$轴交于点$C$,顶点为$P$;抛物线$C_{2}:y = -x^{2} + 2tx + c$,顶点为$Q$.
(1)求抛物线$C_{1}$的表达式及顶点$P$的坐标;
(2)如图(1),连接$PC$,点$D$是抛物线$C_{2}$上一点(点$D$在点$P$右侧),$△PDC$是以$DC$为斜边的直角三角形,若$\tan∠PCD = \frac{3}{4}$,求$t$的值;
(3)如图(2),点$G$为抛物线$C_{2}$与$C_{1}$的异于点$C$的另一个交点,连接$PQ$,$PG$,$QG$,记$△PQG$的面积为$S$,当$S = 1$时,直接写出$t$的值.

答案：
22
(1)方法一:$\because$抛物线$C_{1}:y = x^{2} + bx + c$经过点$A(1,0)$,$B(3,0)$,
$\therefore \begin{cases}0 = 1 + b + c, \\0 = 9 + 3b + c,\end{cases}$
解得$\begin{cases}b = -4, \\c = 3,\end{cases}\therefore$抛物线$C_{1}:y = x^{2} - 4x + 3$.(3分)
$\because y = x^{2} - 4x + 3 = (x - 2)^{2} - 1$,
$\therefore$顶点$P$的坐标为$(2,-1)$.(4分)
方法二:由题意,得抛物线$C_{1}:y = (x - 1)(x - 3)$(点拨:交点式)$= x^{2} - 4x + 3$,且其对称轴为直线$x = \frac{1 + 3}{2} = 2$(点拨:若抛物线过点$(x_{1},y)$,$(x_{2},y)$,则抛物线的对称轴为直线$x = \frac{x_{1} + x_{2}}{2}$).
当$x = 2$时,$y = (x - 1)(x - 3) = -1$,
$\therefore$顶点$P$的坐标为$(2,-1)$.(4分)
(2)由
(1)知抛物线$C_{2}:y = -x^{2} + 2tx + 3$.
将$x = 0$代入$y = x^{2} - 4x + 3$,得$y = 3$,$\therefore C(0,3)$.
如图
(1),过点$P$作$x$轴的平行线,交$y$轴于点$E$,过点$D$作$DF$垂直该平行线于点$F$,

则$\angle CEP = \angle PFD = 90^{\circ}$,$CE = 4$,$EP = 2$,
$\therefore \angle CPE + \angle PCE = 90^{\circ}$.
又$\because \angle CPD = 90^{\circ}$,$\therefore \angle CPE + \angle FPD = 90^{\circ}$,
$\therefore \angle PCE = \angle FPD$,
$\therefore \triangle PCE \sim \triangle DPF$(点拨:“一线三直角”相似模型),
$\therefore \frac{DF}{PE} = \frac{PF}{CE} = \frac{DP}{PC} = \tan \angle PCD = \frac{3}{4}$(关键点:将线段的比例值与已知的正切函数值联系起来),
$\therefore \frac{DF}{2} = \frac{PF}{4} = \frac{3}{4}$,$\therefore DF = \frac{3}{2}$,$PF = 3$,
$\therefore D(2 + \frac{3}{2},-1)$,即$D(\frac{7}{2},-1)$.
将$D(\frac{7}{2},-1)$代入$y = -x^{2} + 2tx + 3$,
得$\frac{1}{2} = -25 + 10t + 3$,解得$t = \frac{9}{4}$.(8分)
(3)$t$的值为$\sqrt{2} - 1$,$-\sqrt{2} - 1$或$-1$.(11分)
解法提示:由抛物线$C_{2}:y = -x^{2} + 2tx + 3$,可得其顶点$Q$的坐标为$(t,t^{2} + 2t + 3)$.
令$x^{2} - 4x + 3 = -x^{2} + 2tx + 3$,
解得$x_{1} = 0$,$x_{2} = t + 2$,$\therefore$点$G(t + 2,t^{2} - 1)$.
易得直线$QG$的表达式为$y = -2x + t^{2} + 2t + 3$.
如图
(2),过点$P$作$x$轴的平行线交$QG$的延长线于点$M$,

点拨:将三边都不与坐标轴平行的三角形的面积转化为有一边与坐标轴平行的两个三角形的面积差
则$M(\frac{1}{2}t^{2} + 2t + 2,-1)$,$\therefore PM = |\frac{1}{2}t^{2} + 2t + 2 - 2| = |\frac{1}{2}t^{2} + 2t|$,
$\therefore S = S_{\triangle PQM} - S_{\triangle PCM} = \frac{1}{2} × PM × 4 = |\frac{1}{2}t^{2} + 2t| × 2 = |t^{2} + 2t|$.
当$S = 1$时,$|t^{2} + 2t| = 1$,$\therefore t^{2} + 2t = 1$或$t^{2} + 2t = -1$,
$\therefore t = \sqrt{2} - 1$,$-\sqrt{2} - 1$或$-1$.
名师辨模型
“一线三直角”模型
1. 图形示例:

2. 证明三角形相似(全等)的关键:利用同(等)角的余角相等这一结论得到一组相等的角.
3. “一线三直角”模型的应用:
(1)图形中已经存在“一线三直角”,直接应用此模型解题;
(2)图形中存在“一线两直角”,补上“一直角”构造此模型;
(3)图形中存在顶点在某条直线上的一个直角,补上“两直角”构造此模型;
(4)对于平面直角坐标系,可以借助$x$轴或$y$轴(也可以借助平行于$x$轴或$y$轴的直线)构造“一线三直角”模型.
名师敲重点
避坑指南
在解决第
(3)问时,由于$t$值不定,故点$M$位于点$P$左侧还是右侧不确定,则在用含$t$的式子表示$PM$的长时,一定不要忘记加绝对值符号,以免造成漏解.