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電子課本網(wǎng) 第102頁(yè)

第102頁(yè)

信息發(fā)布者:
解:?$(1)$?∵拋物線?$C:$??$y=4-(6-x)^2=-(x-6)^2+4$?
∴拋物線的頂點(diǎn)為?$Q(6,$??$4)$?
∴拋物線的對(duì)稱軸為直線?$x=6,$??$y$?的最大值為?$4$?
當(dāng)?$y=3$?時(shí),?$3=-(x-6)^2+4$?
∴?$x=5$?或?$7$?
∵點(diǎn)?$P$?在對(duì)稱軸的右側(cè)
∴?$P(7,$??$3)$?
∴?$a=7$?
?$(2)$?∵平移后的拋物線的解析式為?$y=-(x-3)^2$?
∴平移后的頂點(diǎn)?$Q'(3,$??$0)$?
∵平移前拋物線的頂點(diǎn)?$Q(6,$??$4)$?
∴點(diǎn)?$P'$?移動(dòng)的最短路程?$=QQ'=\sqrt {3^2+4^2}=5$?
解:?$(1)$?∵點(diǎn)?$A$?的坐標(biāo)是?$(-3,$??$0),$?點(diǎn)?$B$?的坐標(biāo)是?$(0,$??$4),$?點(diǎn)?$C$?為?$OB$?中點(diǎn)
∴?$OA=3,$??$OB=4$?
∴?$BC=2$?
將?$△ABC$?繞著點(diǎn)?$B$?逆時(shí)針旋轉(zhuǎn)?$90°$?得到?$△A'BC'$?
∴?$C'(2,$??$4)$?
∵反比例函數(shù)?$y=\frac {k}{x}$?的圖象經(jīng)過點(diǎn)?$C'$?
∴?$k=2×4=8$?
∴該反比例函數(shù)的表達(dá)式為?$y=\frac {8}{x}$?
?$(2)$?作?$A'H⊥y$?軸于?$H$?
∵?$∠AOB=∠A'HB=∠ABA'=90°$?
∴?$∠ABO+∠A'BH=90°,$??$∠ABO+∠BAO=90°$?
∴?$∠BAO=∠A'BH$?
∵?$BA=BA'$?
∴?$△AOB≌△BHA'(\mathrm {AAS})$?
∴?$OA=BH,$??$OB=A'H$?
∵?$OA=3,$??$OB=4$?
∴?$BH=OA=3,$??$A'H=OB=4$?
∴?$OH=1$?
∴?$A'(4,$??$1)$?
設(shè)一次函數(shù)的解析式為?$y=ax+b$?
把?$A(-3,$??$0),$??$A'(4,$??$1)$?代入得?$\begin{cases}{-3a+b=0}\\{4a+b=1}\end{cases},$?解得?$\begin{cases}{a=\dfrac {1}{7}}\\{b=\dfrac {3}{7}}\end{cases}$?
∴該一次函數(shù)的表達(dá)式為?$y=\frac {1}{7}x+\frac {3}{7}$?
解:?$(1)$?以?$C$?為圓心,?$CM$?長(zhǎng)為半徑畫圓,
連接?$CN$?交?$DE$?于?$M_1,$?延長(zhǎng)?$NC$?交圓于?$M_2$?
∵?$△ACB$?是等腰直角三角形,?$N$?是?$AB$?中點(diǎn)
∴?$CN$?平分?$∠ACB,$??$CN=\frac {1}{2}AB=\frac {1}{2}×4=2$?
∵?$△DCE$?是等腰直角三角形,?$M_1$?是?$DE$?中點(diǎn)
∴?$CM_1=\frac {1}{2}DE=\frac {1}{2}×2=1$?
∴?$M、$??$N$?距離的最小值是?$NM_1=CN-CM_1=2-1=1$?
?$M、$??$N$?距離的最大值是?$NM_2=CN+CM_2=2+1=3$?
?$(2)$?連接?$CM,$??$CN,$?作?$NH⊥MC$?交?$MC$?延長(zhǎng)線于?$H$?
∵?$△ACB$?是等腰直角三角形,?$N$?是?$AB$?中點(diǎn)
∴?$CN=\frac {1}{2}AB=2$?
同理:?$CM=\frac {1}{2}DE=1$?
∵?$△CDE$?繞頂點(diǎn)?$C$?逆時(shí)針旋轉(zhuǎn)?$120°$?
∴?$∠MCN=120°$?
∴?$∠NCH=180°-∠MCN=60°$?
∴?$CH=\frac {1}{2}CN=1$?
∴?$NH=\sqrt {3}CH=\sqrt {3}$?
∵?$MH=MC+CH=2$?
∴?$MN=\sqrt {MH^2+NH^2}=\sqrt {7}$?