高考數(shù)學(xué)專(zhuān)題復(fù)習(xí)導(dǎo)練測(cè) 第九章 解析幾何階段測(cè)試(十二)課件 理 新人教A版.ppt
數(shù)學(xué) A(理),45分鐘階段測(cè)試(十二),第九章 平面解析幾何,2,3,4,5,6,7,8,9,10,1,一、選擇題 1.斜率不存在的直線(xiàn)一定是( ) A.過(guò)原點(diǎn)的直線(xiàn) B.垂直于x軸的直線(xiàn) C.垂直于y軸的直線(xiàn) D.垂直于過(guò)原點(diǎn)的直線(xiàn) 解析 斜率不存在,傾斜角為90,故B正確.,B,3,4,5,6,7,8,9,10,1,2,2.已知直線(xiàn)mx4y20與2x5yn0互相垂直,垂足為(1,p),則mnp為( ) A.24 B.20 C.0 D.4,解析 兩直線(xiàn)互相垂直, k1k21,,3,4,5,6,7,8,9,10,1,2,m10.,又垂足為(1,p),,代入直線(xiàn)10x4y20,得p2,,將(1,2)代入直線(xiàn)2x5yn0,得n12,,mnp20.,答案 B,2,4,5,6,7,8,9,10,1,3,3.已知圓的方程為x2y26x8y0.設(shè)該圓過(guò)點(diǎn)(3,5)的最長(zhǎng)弦和最短弦分別為AC和BD,則四邊形ABCD的面積為( ),2,4,5,6,7,8,9,10,1,3,答案 B,2,3,5,6,7,8,9,10,1,4,4.直線(xiàn)l過(guò)點(diǎn)(4,0),且與圓(x1)2(y2)225交于A(yíng),B兩點(diǎn),如果|AB|8,那么直線(xiàn)l的方程為( ) A.5x12y200 B.5x12y200或x40 C.5x12y200 D.5x12y200或x40,2,3,5,6,7,8,9,10,1,4,解析 由題意,得圓心C(1,2),半徑r5, 當(dāng)直線(xiàn)l的斜率不存在時(shí),直線(xiàn)l的方程為x40,,2,3,5,6,7,8,9,10,1,4,即此時(shí)與圓C的交點(diǎn)坐標(biāo)是(4,2)和(4,6),則|AB|8,即x40符合題意;,當(dāng)直線(xiàn)l的斜率存在時(shí),設(shè)直線(xiàn)l的方程為yk(x4),即kxy4k0,,2,3,5,6,7,8,9,10,1,4,即5x12y200.,答案 D,2,3,4,6,7,8,9,10,1,5,5.過(guò)點(diǎn)M(1,2)的直線(xiàn)l與圓C:(x2)2y29交于A(yíng)、B兩點(diǎn),C為圓心,當(dāng)ACB最小時(shí),直線(xiàn)l的方程為( ) A.x1 B.y1 C.xy10 D.x2y30 解析 當(dāng)CMl,即弦長(zhǎng)最短時(shí),ACB最小,,l的方程為:x2y30.,D,2,3,4,5,7,8,9,10,1,6,二、填空題 6.直線(xiàn)l1:2x4y10與直線(xiàn)l2:2x4y30平行,點(diǎn)P是平面直角坐標(biāo)系內(nèi)任一點(diǎn),P到直線(xiàn)l1和l2的距離分別為d1,d2,則d1d2的最小值是_.,2,3,4,5,7,8,9,10,1,6,2,3,4,5,6,8,9,10,1,7,解析 設(shè)所求的圓的方程是(xa)2(yb)2r2,,2,3,4,5,6,8,9,10,1,7,所求的圓與x軸相切,r2b2. ,又所求圓心在直線(xiàn)3xy0上,3ab0. ,聯(lián)立,解得a1,b3,r29或a1,b3,r29.,故所求的圓的方程為(x1)2(y3)29或(x1)2(y3)29.,答案 (x1)2(y3)29或(x1)2(y3)29,2,3,4,5,6,9,10,1,7,8,直線(xiàn)ykx1過(guò)定點(diǎn)(0,1),,在同一坐標(biāo)系中畫(huà)出直線(xiàn)和半圓的草圖,由圖 可知,k的取值范圍是0,1.,0,1,2,3,4,5,6,7,8,10,1,9,三、解答題 9.已知實(shí)數(shù)x、y滿(mǎn)足方程(x3)2(y3)26,求xy的最大值和最小值. 解 設(shè)xyt,則直線(xiàn)yxt與圓(x3)2(y3)26有公共點(diǎn).,2,3,4,5,6,7,8,10,1,9,2,3,4,5,6,7,8,9,1,10,10.已知圓C:x2y22x4y10,O為坐標(biāo)原點(diǎn),動(dòng)點(diǎn)P在圓C外,過(guò)P作圓C的切線(xiàn),設(shè)切點(diǎn)為M. (1)若點(diǎn)P運(yùn)動(dòng)到(1,3)處,求此時(shí)切線(xiàn)l的方程; 解 把圓C的方程化為標(biāo)準(zhǔn)方程為(x1)2(y2)24, 圓心為C(1,2),半徑r2. 當(dāng)l的斜率不存在時(shí),此時(shí)l的方程為x1,,2,3,4,5,6,7,8,9,1,10,C到l的距離d2r,滿(mǎn)足條件. 當(dāng)l的斜率存在時(shí),設(shè)斜率為k, 得l的方程為y3k(x1), 即kxy3k0,,2,3,4,5,6,7,8,9,1,10,即3x4y150. 綜上,滿(mǎn)足條件的切線(xiàn)l的方程為x1或3x4y150.,2,3,4,5,6,7,8,9,1,10,(2)求滿(mǎn)足條件|PM|PO|的點(diǎn)P的軌跡方程. 解 設(shè)P(x,y), 則|PM|2|PC|2|MC|2(x1)2(y2)24, |PO|2x2y2. |PM|PO|. (x1)2(y2)24x2y2,,2,3,4,5,6,7,8,9,1,10,整理,得2x4y10, 點(diǎn)P的軌跡方程為2x4y10.,