26.(1)解:由得點(diǎn)坐標(biāo)為 由得點(diǎn)坐標(biāo)為 ∴··················································································· 由解得∴點(diǎn)的坐標(biāo)為···································· ∴··························································· (2)解:∵點(diǎn)在上且 ∴點(diǎn)坐標(biāo)為······················································································ 又∵點(diǎn)在上且 ∴點(diǎn)坐標(biāo)為······················································································ ∴··········································································· (3)解法一:當(dāng)時(shí).如圖1.矩形與重疊部分為五邊形(時(shí).為四邊形).過作于.則 ∴即∴ ∴ 即··································································· 29. 問題解決 如圖(1).將正方形紙片折疊.使點(diǎn)落在邊上一點(diǎn)(不與點(diǎn).重合).壓平后得到折痕.當(dāng)時(shí).求的值. 類比歸納 在圖(1)中.若則的值等于 ,若則的值等于 ,若(為整數(shù)).則的值等于 .(用含的式子表示) 聯(lián)系拓廣 如圖(2).將矩形紙片折疊.使點(diǎn)落在邊上一點(diǎn)(不與點(diǎn)重合).壓平后得到折痕設(shè)則的值等于 .(用含的式子表示) 查看更多

 

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解:(1)OA=1,OC=2

A點(diǎn)坐標(biāo)為(0,1),C點(diǎn)坐標(biāo)為(2,0)

設(shè)直線AC的解析式為y=kx+b

解得

直線AC的解析式為··················· 2分

(2)

(正確一個(gè)得2分)························· 8分

(3)如圖,設(shè)

點(diǎn)作F

由折疊知

或2··········· 10分

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在直角坐標(biāo)系xOy中,設(shè)點(diǎn)A(0,t),點(diǎn)Q(t,b)(t,b均為非零常數(shù)).平移二次精英家教網(wǎng)函數(shù)y=-tx2的圖象,得到的拋物線F滿足兩個(gè)條件:①頂點(diǎn)為Q;②與x軸相交于B,C兩點(diǎn)(|OB|<|OC|).連接AB.
(1)是否存在這樣的拋物線F,使得|OA|2=|OB|•|OC|?請你作出判斷,并說明理由;
(2)如果AQ∥BC,且tan∠ABO=
32
,求拋物線F對應(yīng)的二次函數(shù)的解析式.

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在直角坐標(biāo)系xOy中,設(shè)點(diǎn)A(0,t),點(diǎn)Q(t,b)(t,b均為非零常數(shù)).平移二次函數(shù)y=-tx2的圖象,得到的拋物線F滿足兩個(gè)條件:①頂點(diǎn)為Q;②與x軸相交于B,C兩點(diǎn)(|OB|<|OC|).連接AB.
(1)是否存在這樣的拋物線F,使得|OA|2=|OB|•|OC|?請你作出判斷,并說明理由;
(2)如果AQ∥BC,且tan∠ABO=數(shù)學(xué)公式,求拋物線F對應(yīng)的二次函數(shù)的解析式.

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在直角坐標(biāo)系xOy中,設(shè)點(diǎn)A(0,t),點(diǎn)Q(t,b)(t,b均為非零常數(shù)),平移二次函數(shù)y=-tx2的圖象,得到的拋物線F滿足兩個(gè)條件:①頂點(diǎn)為Q;②與x軸相交于B,C兩點(diǎn)(|OB|<|OC|),連接AB。

(1)是否存在這樣的拋物線F,使得|OA|2=|OB|·|OC|?請你作出判斷,并說明理由;
(2)如果AQ∥BC,且tan∠ABO=,求拋物線F對應(yīng)的二次函數(shù)的解析式。

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在直角坐標(biāo)系xOy中,設(shè)點(diǎn)A(0,t),點(diǎn)Q(t,b)。平移二次函數(shù)的圖象,得到的拋物線F滿足兩個(gè)條件:①頂點(diǎn)為Q;②與x軸相交于B,C兩點(diǎn)(OB<OC),連結(jié)A,B。

(1)是否存在這樣的拋物線F,使得?請你做出判斷,并說明理由;

(2)如果AQ∥BC,且tan∠ABO=,求拋物線F對應(yīng)的二次函數(shù)的解析式。

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