《2025年全练单元卷九年级数学下册人教版》

25. (10分)已知二次函数$y = ax^{2}-2ax + c(a\gt0)$的图象与$x$轴的负半轴和正半轴分别交于$A$，$B$两点，与$y$轴交于点$C$，它的顶点坐标为$P$，直线$CP$与过点$B$且垂直于$x$轴的直线交于点$D$，且$CP:PD = 2:3$.
(1)求$A$，$B$两点的坐标；
(2)若$\tan\angle PDB = \frac{5}{4}$，求这个二次函数的关系式.
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

答案：
解：
(1)过点$P$作$PE\perp x$轴于点$E$，$\because y = ax^{2}-2ax + c$，$\therefore$该二次函数的对称轴为$x = 1,\therefore OE = 1$，$\because OC// BD,\therefore CP:PD = OE:EB$，$\therefore OE:EB = 2:3$，$\therefore EB=\frac{3}{2},\therefore OB = OE + EB=\frac{5}{2},\therefore B(\frac{5}{2},0)$，$\because A$与$B$关于直线$x = 1$对称，$\therefore A(-\frac{1}{2},0)$；
(2)过点$C$作$CF\perp BD$于点$F$，交$PE$于点$G$，令$x = 1$代入$y = ax^{2}-2ax + c$，$\therefore y = c - a$，令$x = 0$代入$y = ax^{2}-2ax + c$，$\therefore y = c,\therefore PG = a$，$\because CF = OB=\frac{5}{2}$，$\therefore \tan\angle PDB=\frac{CF}{FD},\therefore FD = 2$，

$\because PG// BD,\therefore \triangle CPG\sim\triangle CDF$，$\therefore \frac{PG}{FD}=\frac{CP}{CD}=\frac{2}{5},\therefore PG=\frac{4}{5}$，$\therefore a=\frac{4}{5},\therefore y=\frac{4}{5}x^{2}-\frac{8}{5}x + c$，把$A(-\frac{1}{2},0)$代入$y=\frac{4}{5}x^{2}-\frac{8}{5}x + c$，解得$c = -1$，$\therefore$该二次函数解析式为$y=\frac{4}{5}x^{2}-\frac{8}{5}x - 1$.