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选择性必修三Unit1综合检测题参考答案及部分解析参考答案1-5CBACA6-10ABCAC11-15BACBA16-20ABCCB21-25DBACB26-30DCBBD31-35ABABA36-40DAFCG41-45BACBD46-50ACDBD51-55CABDA56.are57.the58.wasestablished59.opening60.provinces61.original62.itself63.which64.of65.locally写作第一节Onepossibleversion:DearAmy,Alocaltravelagencyislookingforapart-timeEnglish-speakingtourguide,andI'mwritingtosharesomedetailsaboutitincaseyoumaybeinterested.Itsmainresponsibilitiesincludeaccompanyingforeignvisitors,bothindividualsandgroups,tolocaltouristattractions,andgivinginsightsthatcanhelpthemmakethemostoftheexperience.Asmentionedintheadvertisement,nativeEnglishspeakersandthosewhoarefamiliarwithlocalscenicspotsarepreferred.Ithinkyou'reaperfectfitineveryaspect.Ifyouneedanyfurtherinformation,don'thesitatetocontactme.Yours,LiHua第二节Onepossibleversion:Theboy'smouthdroppedopeninexcitement,noddinghishead."I'vecometowelcomeyouintothisroom,theperformertoldhim.Aftertheyoungpatientspoketotheperformerforafewminutes,itwasclearthattheatmosphereintheroomhadswitchedfromstrainedtocalm.SteveaskedSupermantocomeback,andtheperformerpromisedtodosoifhewouldcooperatewiththedoctorsandnurses.ThroughoutSteve'sstayatthehospital,heremainedcooperativeandbrave.Oncehistreatmentwascompleted,Supermanvisitedhimagainaspromised.Withthehelpoftheprogram,theotheryoungpatientsnolongerfearedthehospitaleither.Instead,theyfeltcomfortable,thankstotheirfavoritecharactersvisitingthem.Doctorsandnursesalsohadaneasiertimedealingwiththeirpatients,andoncetheywerereadytoheadhome,theywouldleavethehospitalwithasmileafterwitnessingafarewellperformancefromtheirfavoritesuperheroes.Parentsthankedthehospitalstafffortheirdevotiontothework.Theysaid,totheirchildren,thesearemorethanjustfarewellperformances,butratheracelebrationofabeautifulnewbeginningfortheirkidsastheyleavehospital.部分解析阅读第一节
分析（1）设B（m，n），则$\left\{\begin{array}{l}\frac{n-8}{m-4}=1\\\frac{m+4}{2}+\frac{n+8}{2}=4\end{array}\right.$，解得p值，可得抛物线C的方程；
（2）设E（x1，y1），F（x2，y2），l2：x=sy+t，联立抛物线方程并整理得：y2-16sy-16t=0．结合韦达定理可得$\frac{1}{{D{E^2}}}+\frac{1}{{D{F^2}}}$=$\frac{1}{{t}^{2}}$+$\frac{t-8}{8{t}^{2}（{s}^{2}+1）}$，所以t=8时，存在定点D（8，0），使得$\frac{1}{{D{E^2}}}+\frac{1}{{D{F^2}}}$为定值$\frac{1}{64}$．
解答解：（1）设B（m，n），
则$\left\{\begin{array}{l}\frac{n-8}{m-4}=1\\\frac{m+4}{2}+\frac{n+8}{2}=4\end{array}\right.$
∴$m=-4，n=0，-\frac{p}{2}=-4，p=8$，
所以抛物线C的方程为y2=16x．
（2）设E（x1，y1），F（x2，y2），l2：x=sy+t，
由$\left\{\begin{array}{l}x=sy+t\\{y^2}=16x\end{array}\right.得{y^2}-16sy-16t=0$．
其中△=（16s）2+64t＞0，则y1+y2=16s，y1y2=-16t，
$\frac{1}{{D{E^2}}}+\frac{1}{{D{F^2}}}$=$\frac{1}{（{x}_{1}-t）^{2}+{y}_{1}^{2}}$+$\frac{1}{{（{x}_{2}-t）}^{2}+{y}_{2}^{2}}$=$\frac{1}{（{s}^{2}+1）{y}_{1}^{2}}$+$\frac{1}{（{s}^{2}+1）{y}_{2}^{2}}$=$\frac{{y}_{1}^{2}+{y}_{1}^{2}}{（{s}^{2}+1）{y}_{1}^{2}{y}_{2}^{2}}$=$\frac{{（{y}_{1}^{\;}+{y}_{2}^{\;}）}^{2}-{2y}_{1}{y}_{2}}{（{s}^{2}+1）{y}_{1}^{2}{y}_{2}^{2}}$=$\frac{8{s}^{2}+t}{8{t}^{2}（{s}^{2}+1）}$=$\frac{1}{{t}^{2}}$+$\frac{t-8}{8{t}^{2}（{s}^{2}+1）}$，
所以t=8时，存在定点D（8，0），使得$\frac{1}{{D{E^2}}}+\frac{1}{{D{F^2}}}$为定值$\frac{1}{64}$．
点评本题考查的知识点是抛物线的简单性质，直线与圆锥曲线的位置关系，难度中档．
