（魯京津瓊專(zhuān)用）2020版高考數(shù)學(xué)復(fù)習(xí)第九章平面解析幾何階段自測(cè)卷（六）課件.pptx

階段自測(cè)卷(六),第九章 平面解析幾何,一、選擇題(本大題共12小題，每小題5分，共60分) 1.(2019四川診斷)拋物線(xiàn)y24x的焦點(diǎn)坐標(biāo)是,由拋物線(xiàn)y24x得2p4，解得 p2， 則焦點(diǎn)坐標(biāo)為(1,0)，故選C.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,2.(2019撫州七校聯(lián)考)過(guò)點(diǎn)(2,1)且與直線(xiàn)3x2y0垂直的直線(xiàn)方程為 A.2x3y10 B.2x3y70 C.3x2y40 D.3x2y80,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 設(shè)要求的直線(xiàn)方程為2x3ym0， 把點(diǎn)(2,1)代入可得43m0，解得m7. 可得要求的直線(xiàn)方程為2x3y70，故選B.,3.(2019陜西四校聯(lián)考)直線(xiàn)axby0與圓x2y2axby0的位置關(guān)系是 A.相交 B.相切 C.相離 D.不能確定,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,圓與直線(xiàn)的位置關(guān)系是相切.故選B.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 右焦點(diǎn)F到漸近線(xiàn)的距離為2，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,橢圓C的長(zhǎng)軸長(zhǎng)與焦距之和為6，即2a2c6，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 由條件，得|OP|22ab，又P為雙曲線(xiàn)上一點(diǎn)， 從而|OP|a，2aba2，2ba，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 由題意，過(guò)原點(diǎn)O且傾斜角為30的直線(xiàn)l與橢圓C的一個(gè)交點(diǎn)為A， 且AF1AF2，且 2，則可知|OA|c，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解得c24，且c2a2b2， 所以a26，b22，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,即點(diǎn)M(x，y)到拋物線(xiàn)y24x的準(zhǔn)線(xiàn)x1的距離，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,10.(2019河北衡水中學(xué)調(diào)研)已知y24x的準(zhǔn)線(xiàn)交x軸于點(diǎn)Q，焦點(diǎn)為F，過(guò)Q且斜率大于0的直線(xiàn)交y24x于A，B，兩點(diǎn)AFB60，則|AB|等于,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,|AF|x11，|BF|x21， 代入余弦定理|AB|2|AF|2|BF|22|AF|BF|cos 60，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,11.(2019成都七中診斷)設(shè)拋物線(xiàn)C：y212x的焦點(diǎn)為F，準(zhǔn)線(xiàn)為l，點(diǎn)M在C上，點(diǎn)N在l上，且 (0)，若|MF|4，則等于,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,又|MF|4，|MM|4， 又|FF|6，,故選D.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 根據(jù)題意，可知|PF1|PF2|2a， |PF1|PF2|2m， 解得|PF1|am，|PF2|am， 根據(jù)余弦定理，可知,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,二、填空題(本大題共4小題，每小題5分，共20分),1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,14.(2019南昌八一中學(xué)、洪都中學(xué)聯(lián)考)若F1，F(xiàn)2是橢圓 1的左、右焦點(diǎn)，點(diǎn)P在橢圓上運(yùn)動(dòng)，則|PF1|PF2|的最大值是_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,當(dāng)且僅當(dāng)|PF1|PF2|時(shí)取等號(hào)， 所以|PF1|PF2|的最大值為5.,21,22,5,15.(2018蘭州調(diào)研)點(diǎn)P在圓C1：x2y28x4y110上，點(diǎn)Q在圓C2：x2y24x2y10上，則|PQ|的最小值是_.,解析 把圓C1、圓C2的方程都化成標(biāo)準(zhǔn)形式，得 (x4)2(y2)29，(x2)2(y1)24. 圓C1的圓心坐標(biāo)是(4,2)，半徑是3； 圓C2的圓心坐標(biāo)是(2，1)，半徑是2.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,16.(2019廣東六校聯(lián)考)已知直線(xiàn)l：ykxt與圓C1：x2(y1)22相交于A，B兩點(diǎn)，且C1AB的面積取得最大值，又直線(xiàn)l與拋物線(xiàn)C2：x22y相交于不同的兩點(diǎn)M，N，則實(shí)數(shù)t的取值范圍是_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,(，4)(0，),當(dāng)角度為直角時(shí)面積最大，此時(shí)C1AB為等腰直角三角形， 則圓心到直線(xiàn)的距離為d1，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,直線(xiàn)l與拋物線(xiàn)C2：x22y相交于不同的兩點(diǎn)M，N， 聯(lián)立直線(xiàn)和拋物線(xiàn)方程得到x22kx2t0 ， 只需要此方程有兩個(gè)不等根即可，4k28t4t216t0 ， 解得t的取值范圍為(，4)(0，).,三、解答題(本大題共70分) 17.(10分)(2018重慶朝陽(yáng)中學(xué)月考)已知直線(xiàn)l1：ax2y60，直線(xiàn)l2：x(a1)ya210. (1)求a為何值時(shí)，l1l2；,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解得a1或a2(舍去)， 當(dāng)a1時(shí)，l1l2.,(2)求a為何值時(shí)，l1l2.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解 l1l2，a12(a1)0，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,18.(12分)已知圓C：(x1)2(y2)225，直線(xiàn)l：(2m1)x(m1)y7m40. (1)證明：對(duì)任意實(shí)數(shù)m，直線(xiàn)l恒過(guò)定點(diǎn)且與圓C交于兩個(gè)不同點(diǎn)；,證明 直線(xiàn)l：(2m1)x(m1)y7m40可化為m(2xy7)(xy4)0，,所以直線(xiàn)l恒過(guò)點(diǎn)P(3,1)，而點(diǎn)P(3,1)在圓C內(nèi)， 所以對(duì)任意實(shí)數(shù)m，直線(xiàn)l恒過(guò)點(diǎn)P(3,1)且與圓C交于兩個(gè)不同點(diǎn).,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,(2)求直線(xiàn)l被圓C截得的弦長(zhǎng)最小時(shí)的方程.,解 由(1)得，直線(xiàn)l恒過(guò)圓C內(nèi)的定點(diǎn)P(3,1)， 設(shè)過(guò)點(diǎn)P的弦長(zhǎng)為a，過(guò)圓心C向直線(xiàn)l作垂線(xiàn)，垂足為弦的中點(diǎn)H，,當(dāng)且僅當(dāng)H與P重合時(shí)取等號(hào)， 此時(shí)弦所在的直線(xiàn)與直線(xiàn)CP垂直，又過(guò)點(diǎn)P(3,1)， 所以，當(dāng)直線(xiàn)l被圓C截得的弦長(zhǎng)最小時(shí)，弦所在的直線(xiàn)方程為2xy50.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,b2a2c24,(2)求PAB的面積.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解 設(shè)直線(xiàn)l的方程為yxm，代入橢圓方程得 4x26mx3m2120， (*) 設(shè)A(x1，y1)，B(x2，y2)，AB的中點(diǎn)為E(x0，y0),因?yàn)锳B是等腰PAB的底邊，所以PEAB.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,此時(shí)方程(*)為4x212x0.,此時(shí)，點(diǎn)P(3,2)到直線(xiàn)AB：xy20的距離,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20.(12分)(2019四川診斷)已知橢圓C： 1(ab0)的左焦點(diǎn)F(2,0)，上頂點(diǎn)B(0,2). (1)求橢圓C的方程；,解 由題意可得c2，b2，,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,(2)若直線(xiàn)yxm與橢圓C交于不同兩點(diǎn)M，N，且線(xiàn)段MN的中點(diǎn)G在圓x2y21上，求m的值.,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,解 設(shè)點(diǎn)M，N的坐標(biāo)分別為(x1，y1)，(x2，y2)， 線(xiàn)段MN的中點(diǎn)G(x0，y0)，,21,22,因?yàn)辄c(diǎn)G(x0，y0)在圓x2y21上，,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,(2)過(guò)左焦點(diǎn)F的直線(xiàn)與橢圓分別交于A，B兩點(diǎn)，若OAB(O為直角坐標(biāo)原點(diǎn))的面積為 ，求直線(xiàn)AB的方程.,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,當(dāng)直線(xiàn)AB與x軸不垂直時(shí)， 設(shè)直線(xiàn)AB的方程為yk(x1)，,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,22.(12分)(2019新鄉(xiāng)模擬)如圖，已知橢圓C： 1(ab0)的左、右焦點(diǎn)分別為F1，F(xiàn)2，|F1F2|2，過(guò)點(diǎn)F1的直線(xiàn)與橢圓C交于A，B兩點(diǎn)，延長(zhǎng)BF2交橢圓C于點(diǎn)M，ABF2的周長(zhǎng)為8. (1)求橢圓C的離心率及方程；,21,22,解 由題意可知，|F1F2|2c2，則c1， 又ABF2的周長(zhǎng)為8，所以4a8，即a2，,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,(2)試問(wèn)：是否存在定點(diǎn)P(x0,0)，使得 為定