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典例1如图,点A,E(2,1)在反比例函数y=$\frac{k} {x}$(x＞0)的图像上,点B在反比例函数y=$\frac{6} {x}$(x＜0)的图像上,分别过点A,B作y轴的垂线,垂足分别为C,D,AC=BD,连接AB交y轴于点F.
(1)k=________;
(2)设点A的横坐标为a,点F的纵坐标为m,求证:am=−2;
(3)连接CE,DE,若∠CED=90°,则点A的坐标为________.
[思路点拨](1)将点E的坐标代入反比例函数的表达式即可求得k的值.(2)可证△ACF≌△BDF,得CF=DF,用含a,m的式子表示出线段长,列式即可证得结论.(3)设点A的坐标为(a,$\frac{2} {a}$),则可得点C的坐标为(0,$\frac{2} {a}$),点D的坐标为(0,−$\frac{6} {a}$),根据勾股定理求出a的值,便可求得点A的坐标.
(1)k=________;
(2)设点A的横坐标为a,点F的纵坐标为m,求证:am=−2;
(3)连接CE,DE,若∠CED=90°,则点A的坐标为________.
[思路点拨](1)将点E的坐标代入反比例函数的表达式即可求得k的值.(2)可证△ACF≌△BDF,得CF=DF,用含a,m的式子表示出线段长,列式即可证得结论.(3)设点A的坐标为(a,$\frac{2} {a}$),则可得点C的坐标为(0,$\frac{2} {a}$),点D的坐标为(0,−$\frac{6} {a}$),根据勾股定理求出a的值,便可求得点A的坐标.
答案
专题二 与反比例函数有关的综合题
典例1(1)2.
(2)由(1)知,反比例函数的表达式为$y = \frac{2}{x}$。
$\because AC\perp y$轴，$BD\perp y$轴，$\therefore \angle ACF = \angle BDF = 90^{\circ}$。在$\triangle ACF$和$\triangle BDF$中，$\begin{cases} \angle AFC = \angle BFD, \\ \angle ACF = \angle BDF, \\ AC = BD, \end{cases}$ $\therefore \triangle ACF\cong \triangle BDF$。$\therefore CF = DF$。
$\because$点$A$的横坐标为$a$，点$A$在反比例函数$y = \frac{2}{x}$的图像上，$\therefore$点$A$的坐标为$(a,\frac{2}{a})$。$\therefore$易得点$C$的坐标为$(0,\frac{2}{a})$，点$B$的坐标为$(-a,-\frac{6}{a})$，则点$D$的坐标为$(0,-\frac{6}{a})$。$\therefore AC = BD = a$，$OC = \frac{2}{a}$，$OD = \frac{6}{a}$。
$\because$点$F$的纵坐标为$m$，$\therefore OF = -m$。$\because CF = DF$，$\therefore OC + OF = OD - OF$，即$\frac{2}{a}-m=\frac{6}{a}+m$。整理，得$am = -2$。
(3)设点$A$的坐标为$(a,\frac{2}{a})$，则点$C$的坐标为$(0,\frac{2}{a})$，点$D$的坐标为$(0,-\frac{6}{a})$。
$\because E(2,1)$，$\angle CED = 90^{\circ}$，$\therefore CE^{2}+DE^{2}=CD^{2}$，即$2^{2}+(1 - \frac{2}{a})^{2}+2^{2}+(1+\frac{6}{a})^{2}=(\frac{2}{a}+\frac{6}{a})^{2}$，解得$a_{1} = -2$（不合题意，舍去），$a_{2}=\frac{6}{5}$。$\therefore$点$A$的坐标为$(\frac{6}{5},\frac{5}{3})$。
典例1(1)2.
(2)由(1)知,反比例函数的表达式为$y = \frac{2}{x}$。
$\because AC\perp y$轴，$BD\perp y$轴，$\therefore \angle ACF = \angle BDF = 90^{\circ}$。在$\triangle ACF$和$\triangle BDF$中，$\begin{cases} \angle AFC = \angle BFD, \\ \angle ACF = \angle BDF, \\ AC = BD, \end{cases}$ $\therefore \triangle ACF\cong \triangle BDF$。$\therefore CF = DF$。
$\because$点$A$的横坐标为$a$，点$A$在反比例函数$y = \frac{2}{x}$的图像上，$\therefore$点$A$的坐标为$(a,\frac{2}{a})$。$\therefore$易得点$C$的坐标为$(0,\frac{2}{a})$，点$B$的坐标为$(-a,-\frac{6}{a})$，则点$D$的坐标为$(0,-\frac{6}{a})$。$\therefore AC = BD = a$，$OC = \frac{2}{a}$，$OD = \frac{6}{a}$。
$\because$点$F$的纵坐标为$m$，$\therefore OF = -m$。$\because CF = DF$，$\therefore OC + OF = OD - OF$，即$\frac{2}{a}-m=\frac{6}{a}+m$。整理，得$am = -2$。
(3)设点$A$的坐标为$(a,\frac{2}{a})$，则点$C$的坐标为$(0,\frac{2}{a})$，点$D$的坐标为$(0,-\frac{6}{a})$。
$\because E(2,1)$，$\angle CED = 90^{\circ}$，$\therefore CE^{2}+DE^{2}=CD^{2}$，即$2^{2}+(1 - \frac{2}{a})^{2}+2^{2}+(1+\frac{6}{a})^{2}=(\frac{2}{a}+\frac{6}{a})^{2}$，解得$a_{1} = -2$（不合题意，舍去），$a_{2}=\frac{6}{5}$。$\therefore$点$A$的坐标为$(\frac{6}{5},\frac{5}{3})$。
