1. 如图,在平面直角坐标系中,矩形ABCD的顶点A,D分别在x轴、y轴上,对角线BD//x轴,反比例函数$y=\frac{k}{x}(k＞0,x＞0)$的图象经过矩形对角线的交点E.若点A(2,0),D(0,4),则k的值为 ()

A.16
B.20
C.32
D.40

1. B 解析:设$B(x,4)$,由题易知E为BD中点,$∠ DAB=90°$,
$\therefore E(\dfrac{1}{2}x,4).\because AD^{2}+AB^{2}=BD^{2},\therefore 2^{2}+4^{2}+(x-2)^{2}+4^{2}=x^{2}$,
解得$x=10,\therefore E(5,4),\therefore k=20$.

2.（陕西中考）如图，在矩形OABC和正方形CDEF中，点A在y轴正半轴上，点C，F均在x轴正半轴上，点D在边BC上，BC=2CD，AB=3.若点B，E在同一个反比例函数的图象上，则这个反比例函数的解析式是.

2. $y=\frac{18}{x}$ 解析:$\because$ 四边形OABC是矩形,$\therefore OC=AB=3.\because$ 四边形CDEF是正方形,$\therefore CD=CF=EF.\because BC=2CD,\therefore$ 设$CD=m,BC=2m,\therefore B(3,2m),E(3+m,m)$.设反比例函数的解析式为$y=\dfrac{k}{x},\therefore 3× 2m=(3+m)· m$,解得$m=3$或$m=0$(不合题意,舍去),$\therefore B(3,6),\therefore k=3× 6=18,\therefore$ 这个反比例函数的解析式是$y=\dfrac{18}{x}$.

3. (达州中考)如图,一次函数$y=2x$与反比例函数$y=\dfrac{2}{x}$的图象相交于A,B两点,以AB为边作等边三角形ABC,若反比例函数$y=\dfrac{k}{x}$的图象过点C,则k的值为.
(第3题)

3. -6 解析:由题意,联立$\begin{cases} y=2x, \\ y=\dfrac{2}{x}, \end{cases}$解得$\begin{cases} x=1, \\ y=2 \end{cases}$或$\begin{cases} x=-1, \\ y=-2. \end{cases}$
$\therefore A(1,2),B(-1,-2)$.如图,过点C作$CE⊥ x$轴于点E,过点A作$AD⊥ x$轴于点D,连接OC.$\because OD=1,AD=2,\therefore OA=\sqrt{1^{2}+2^{2}}=\sqrt{5},\therefore AO=BO=\sqrt{5}.\because △ ABC$是等边三角形,
$\therefore CO⊥ AB,∠ ACO=∠ BCO=\dfrac{1}{2}∠ ACB=30°,\therefore AC=2OA=2\sqrt{5},\therefore OC=\sqrt{AC^{2}-OA^{2}}=\sqrt{15}$.设点$C(a,b)$,则$a^{2}+b^{2}=15,AC^{2}=(a-1)^{2}+(b-2)^{2}=(2\sqrt{5})^{2}$,解得$a=-2b$,代入$a^{2}+b^{2}=15$,得$b=\pm\sqrt{3},\therefore C(2\sqrt{3},-\sqrt{3})$或$(-2\sqrt{3},\sqrt{3})$.将点C的坐标代入$y=\dfrac{k}{x}$,得$k=-6$.

4.（福建中考）如图，菱形ABCD的顶点A在函数$y=\frac{3}{x}(x＞0)$的图象上，函数$y=\frac{k}{x}(k＞3,x＞0)$的图象关于直线AC对称，且经过B，D两点，若AB=2，∠BAD=30°，则k=。

4. $6+2\sqrt{3}$ 解析:如图,连接OC,AC,过点A作$AE⊥ x$轴于点E,延长DA与x轴交于点F,过点D作$DG⊥ x$轴于点G,易知O,A,C三点在同一直线上,且$∠ COE=45°,\therefore OE=AE$.设$OE=AE=a$,则$A(a,a)$.将$A(a,a)$代入$y=\dfrac{3}{x}$,得$a=\sqrt{3}$,$\therefore AE=OE=\sqrt{3}.\because ∠ BAD=30°,\therefore ∠ OAF=∠ CAD=15°$,$\therefore ∠ EAF=30°,\therefore AF=2,EF=1$.又由题知$AD=AB=2$,$\therefore A$为DF的中点,故$EF=EG=1,DG=2AE=2\sqrt{3}$,$\therefore OG=OE+EG=\sqrt{3}+1$,$\therefore D(\sqrt{3}+1,2\sqrt{3})$.将$D(\sqrt{3}+1,2\sqrt{3})$代入$y=\dfrac{k}{x}(k＞3,x＞0)$中,可得$k=(\sqrt{3}+1)× 2\sqrt{3}=6+2\sqrt{3}$.

5. (2025·福建中考)如图,平面直角坐标系$xOy$中,$□ OABC$的顶点$A$在$y$轴正半轴上,反比例函数$y=\frac{k}{x}(k＞0,x＞0)$的图象经过$AB$的中点$D$,与边$BC$相交于点$E$,且反比例函数$y=-\frac{k}{x}(k＞0,x＞0)$的图象经过点$C$,连接$DE$,则$△ BDE$与$□ OABC$的面积比是。

5. $\frac{1}{20}$ 解析:$\because$ 反比例函数$y=\dfrac{k}{x}(k＞0,x＞0)$的图象经过AE的中点D,$\therefore$ 设$D(a,\dfrac{k}{a})(a＞0).\because □ OABC$的顶点A在y轴正半轴上,$\therefore OA// BC$,点A的横坐标为0.$\because \dfrac{x_{B}+x_{A}}{2}=a$,即$\dfrac{x_{B}+0}{2}=a$,$\therefore x_{B}=2a$,$\therefore$ 点B的横坐标为$2a$,$\therefore$ 点E,C的横坐标均为$2a$.$\because$ 点E在反比例函数$y=\dfrac{k}{x}(k＞0,x＞0)$的图象上,$\therefore y=\dfrac{k}{2a}$,即$E(2a,\dfrac{k}{2a}).\because$ 点C在反比例函数$y=-\dfrac{k}{x}(k＞0,x＞0)$的图象上,$\therefore y=-\dfrac{k}{2a}$,即$C(2a,-\dfrac{k}{2a})$,设$B(2a,m)$,则$OA=BC=m-(-\dfrac{k}{2a})=m+\dfrac{k}{2a}$,$\therefore A(0,m+\dfrac{k}{2a})$,且AB的中点$D(a,\dfrac{k}{a})$,$\therefore \dfrac{y_{A}+y_{B}}{2}=\dfrac{m+\dfrac{k}{2a}+m}{2}=\dfrac{k}{a}$,解得$m=\dfrac{3k}{4a}$,$\therefore B(2a,\dfrac{3k}{4a})$,$OA=m+\dfrac{k}{2a}=\dfrac{3k}{4a}+\dfrac{k}{2a}=\dfrac{5k}{4a}$,$\therefore BE=\dfrac{3k}{4a}-\dfrac{k}{2a}=\dfrac{k}{4a}$,$\therefore S_{△ BDE}=\dfrac{1}{2}BE· |x_{B}-x_{D}|=\dfrac{1}{2}×\dfrac{k}{4a}×(2a-a)=\dfrac{k}{8}$,$S_{□ OABC}=OA· x_{B}=\dfrac{5k}{4a}×2a=\dfrac{5k}{2}$,$\therefore △ BDE$与$□ OABC$的面积比是$\dfrac{k}{8}:\dfrac{5k}{2}=\dfrac{1}{20}$.

6. (2026·白银期末)已知,矩形OCBA在平面直角坐标系中的位置如图所示,点C在x轴的正半轴上,点A在y轴的正半轴上,已知点B坐标为(3,6),反比例函数$y=\frac{m}{x}(x＞0)$的图象经过AB的中点D,且与BC交于点E,顺次连接O,D,E.
(1)求m的值及点E的坐标.
(2)点M为y轴正半轴上一点,若$△ MBO$的面积等于$△ ODE$的面积,求点M的坐标.
(3)平面直角坐标系中是否存在一点N,使得O,D,E,N四点连接构成平行四边形?若存在,请直接写出N的坐标;若不存在,请说明理由.

6. (1)$\because$ 四边形OCBA为矩形,$\therefore AB// OC,BC// OA.\because$ 点B坐标为(3,6),D为AB的中点,$\therefore D(\dfrac{3}{2},6).\because$ 反比例函数$y=\dfrac{m}{x}(x＞0)$的图象经过AB的中点D,$\therefore m=\dfrac{3}{2}×6=9$,$\therefore$ 反比例函数解析式为$y=\dfrac{9}{x}$,当$x=3$时,$y=\dfrac{9}{3}=3$,$\therefore$ 点E的坐标为(3,3).
(2)$\because$ 点B坐标为(3,6),点D坐标为$(\dfrac{3}{2},6)$,点E坐标为(3,3),$\therefore S_{△ ODE}=3×6-\dfrac{1}{2}×\dfrac{3}{2}×6-\dfrac{1}{2}×3×3-\dfrac{1}{2}×\dfrac{3}{2}×3=18-\dfrac{9}{2}-\dfrac{9}{2}-\dfrac{9}{4}=\dfrac{27}{4}$.$\because$ 点M为y轴正半轴上一点,$\therefore$ 设点M的坐标为$(0,a)$.又$\because △ MBO$的面积等于$△ ODE$的面积,$\therefore \dfrac{3a}{2}=\dfrac{27}{4}$,解得$a=\dfrac{9}{2}$,$\therefore$ 点M的坐标为$(0,\dfrac{9}{2})$.
(3)存在.点N的坐标为$(-\dfrac{3}{2},3)$或$(\dfrac{3}{2},-3)$或$(\dfrac{9}{2},9)$.
解析:①当OD为对角线时,如图有四边形OEDN₁为平行四边形,$\therefore DE=ON_{1},DE// ON_{1}$,延长$N_{1}D$至$N_{3}$,使得$N_{3}D=N_{1}D$,$\therefore ∠ N_{3}DE=∠ N_{3}N_{1}O$.过点$N_{1}$作$N_{1}P⊥ y$轴于点P,有$AB// N_{1}P$,$\therefore ∠ N_{3}DB=∠ N_{3}N_{1}P$,$\therefore ∠ N_{3}DE-∠ N_{3}DB=∠ N_{3}N_{1}O-∠ N_{3}N_{1}P$,即$∠ BDE=∠ PN_{1}O$.又$\because ∠ DBE=∠ N_{1}PO=90°$,$DE=N_{1}O$,$\therefore △ DBE≌ △ N_{1}PO$ (AAS),$\therefore N_{1}P=DB=\dfrac{3}{2}$,$OP=BE=3$,$\therefore N_{1}(-\dfrac{3}{2},3)$.

②当OE为对角线时,如图有四边形$ON_{2}ED$为平行四边形,$\therefore ON_{2}=DE=ON_{1}$.过点$N_{2}$作$N_{2}Q⊥ y$轴于点Q,$∠ N_{1}PO=∠ OQN_{2}=90°$,$∠ N_{1}OP=∠ N_{2}OQ$,$\therefore △ N_{1}OP≌ △ N_{2}OQ$ (AAS),$\therefore N_{2}Q=N_{1}P=\dfrac{3}{2}$,$OQ=OP=3$,$\therefore N_{2}(\dfrac{3}{2},-3)$.
③当DE为对角线时,如图有四边形$OEN_{3}D$为平行四边形,过点$N_{3}$作$N_{3}S$垂直于CB的延长线于点S,延长BC,$QN_{2}$交于点R,则$R(3,-3)$.由②同理可证$△ N_{3}SE≌ △ N_{2}RE$ (AAS),$\therefore N_{3}S=N_{2}R=3-\dfrac{3}{2}=\dfrac{3}{2}$,$SE=RE=3-(-3)=6$,$\therefore N_{3}(\dfrac{9}{2},9)$.
综上所述,点N的坐标为$(-\dfrac{3}{2},3)$或$(\dfrac{3}{2},-3)$或$(\dfrac{9}{2},9)$.