2025優(yōu)化設(shè)計(jì)一輪課時(shí)規(guī)范練69　直線(xiàn)與橢圓

課時(shí)規(guī)范練69直線(xiàn)與橢圓一、基礎(chǔ)鞏固練1.直線(xiàn)y=x+1與橢圓x2+y22=1的位置關(guān)系是(A.相離 B.相切C.相交 D.無(wú)法確定2.已知直線(xiàn)y=-x+1與橢圓x2a2+y2b2=1(a>b>0)相交于A,A.223 B.423 C.23.(2024·吉林長(zhǎng)春模擬)已知橢圓x216+y29A.8x-6y-7=0 B.3x+4y=0C.3x+4y-12=0 D.6x+8y-25=04.已知橢圓x22+y2=1與直線(xiàn)y=x+m交于A,B兩點(diǎn),且|AB|=423,則實(shí)數(shù)A.±1 B.±12 C.2 D.±5.(2024·遼寧遼陽(yáng)模擬)已知直線(xiàn)y=-12x+2與橢圓C:x2a2+y2b2=1(a>b>0)交于A,BA.32 B.12 C.34 6.(2023·新高考Ⅱ,5)已知橢圓C:x23+y2=1的左、右焦點(diǎn)分別為F1,F2,直線(xiàn)y=x+m與C交于A,B兩點(diǎn),若△F1AB面積是△F2AB面積的2倍,則m=(A.23 B.23 C.-23 D7.已知橢圓x29+y2=1,過(guò)左焦點(diǎn)F作傾斜角為π6的直線(xiàn)交橢圓于A,B兩點(diǎn),則弦AB的長(zhǎng)為8.已知橢圓C:x2a2+y2b2=1(a>b>0)的右焦點(diǎn)為F,過(guò)點(diǎn)F的直線(xiàn)與橢圓交于A,B兩點(diǎn).若AB的中點(diǎn)為P19.(2024·河南鄭州模擬)直線(xiàn)l:x+y-1=0與橢圓C:x24+y22=1交于A,B兩點(diǎn),橢圓的右頂點(diǎn)為P,則10.已知橢圓C:x2a2+y2b2(1)求橢圓C的標(biāo)準(zhǔn)方程;(2)過(guò)橢圓C的右焦點(diǎn)且傾斜角為135°的直線(xiàn)l交橢圓C于M,N兩點(diǎn),求|MN|的值.二、綜合提升練11.(2024·江西九江模擬)已知直線(xiàn)l1:y=2x+2過(guò)橢圓C:x2a2+y2b2=1(a>b>0)的一個(gè)焦點(diǎn),與C交于A,B兩點(diǎn),與l1平行的直線(xiàn)l2與C交于M,N兩點(diǎn).若AB的中點(diǎn)為P,MN的中點(diǎn)為Q,且PQA.x24+y23=1C.x29+y28=112.橢圓C:ax2+by2=1與直線(xiàn)x+y=1相交于A,B兩點(diǎn),D是AB的中點(diǎn),O為坐標(biāo)原點(diǎn),OD的斜率為12,則橢圓C的離心率為.13.直線(xiàn)x+y-1=0截橢圓x24+y23=1所得弦的中點(diǎn)M14.(2024·廣東云浮模擬)已知橢圓E:x2a2+y2b2=(1)求橢圓E的方程;(2)斜率為1的直線(xiàn)l與橢圓E交于A,B兩點(diǎn),以AB為底邊作等腰三角形,頂點(diǎn)為C(-3,2),求△PAB的面積.15.已知橢圓C:x2a2+y2b2=1(a>b>0)過(guò)點(diǎn)(1,62),直線(xiàn)l:y=x+m與橢圓C交于A,B兩點(diǎn),且線(xiàn)段AB的中點(diǎn)為M,(1)求橢圓C的標(biāo)準(zhǔn)方程.(2)當(dāng)m=1時(shí),橢圓C上是否存在P,Q兩點(diǎn),使得P,Q關(guān)于直線(xiàn)l對(duì)稱(chēng)?若存在,求出P,Q的坐標(biāo);若不存在,請(qǐng)說(shuō)明理由.

課時(shí)規(guī)范練69直線(xiàn)與橢圓1.C解析聯(lián)立y=x+1,x2+y22=1,得3x2+2x-1=0,則Δ=222.B解析由條件知c=1,e=ca=22,所以a=2,b=1,橢圓方程為x2聯(lián)立直線(xiàn)方程與橢圓方程可得交點(diǎn)坐標(biāo)為(0,1),43,-133.C解析設(shè)弦的兩個(gè)端點(diǎn)分別為A(x1,y1),B(x2,y2),則x1216+y129=1,即(x1-x2)(x1+x2)16=-(y1+y2)(y1-y2)9,易知故以點(diǎn)2,32為中點(diǎn)的弦所在的直線(xiàn)方程為y-32=-整理得3x+4y-12=0.4.A解析由x22+y2=1,y=x+m,消去y并整理,得3設(shè)A(x1,y1),B(x2,y2),則x1+x2=-4m3,x1x2由題意,得|AB|=2(x1+x5.A解析設(shè)A(x1,y1),B(x2,y2),則x12a2+y12b2=1,x22由題意可得x1+x2=4,y1+y2=2,y1-y2x1-x2=-12,則-4b6.C解析如圖所示,橢圓x23+y2=1的左、右焦點(diǎn)分別為F1(-2,0),F2(設(shè)點(diǎn)F1,F2到直線(xiàn)y=x+m的距離分別為d1,d2,由點(diǎn)到直線(xiàn)的距離公式可知d1=|-2+m|2由x23+y2=1,y=x+m,消去y可得4∵y=x+m與橢圓C交于A,B兩點(diǎn),∴Δ=36m2-16(3m2-3)>0,即-2<m<2.∵△F1AB的面積是△F2AB的兩倍,∴有12·|AB|·d1=2×12·|AB|·d2,即d1=2d2,兩邊平方整理,得3m2+102m+6=0,解得m=-23或m=-3又-2<m<2,∴m=-23.故選7.2解析在橢圓x29+y2=1中,a=3,b=1,則c=a2-b2=22,故點(diǎn)F設(shè)點(diǎn)A(x1,y1),B(x2,y2),由題意可知,直線(xiàn)AB的方程為y=33(x+22),即x=3y-22聯(lián)立x=3y-22,x2+9y2=9,可得12y2-46y-1=0,Δ=16×6+4×12=144>0,則所以|AB|=1+3·(y1+y8.2x29+4y29=1解析∵12c-1=1,∴c=32.令則x12a2+∴(x1+x2)(x1-x2)a2+(y1+y2)(y1-y2)9.103解析聯(lián)立x24+y22=1,x+y-1=0,得3x2-4x-2=0.設(shè)A(x1,y1),B(x2,y所以|AB|=1+1易知點(diǎn)P(2,0),故點(diǎn)P到直線(xiàn)l:x+y-1=0的距離d=|所以△ABP的面積S=12|AB|·d=10.解(1)由題得1解得a∴橢圓C的標(biāo)準(zhǔn)方程為x24+(2)由(1)知橢圓C的右焦點(diǎn)坐標(biāo)為(1,0).又直線(xiàn)l的斜率k=-1,∴直線(xiàn)l的方程為x+y-1=0.設(shè)M(x1,y1),N(x2,y2),聯(lián)立x24+y23=1,x+∴x1+x2=87,x1x2=-8∴|MN|=1+k2|x1-x211.C解析設(shè)A(x1,y1),B(x2,y2),P(x0,y0),易知x1≠x2,則x兩式作差得(x1+所以y1-若O為坐標(biāo)原點(diǎn),則kOP=-b22a2,同理kOQ=-b22a2,即kOP=kPQ=-b22a2=-49,所以b2a2=89.又l1過(guò)點(diǎn)(-1,0),所以C的方程為x29+12.22解析設(shè)A(x1,y1),B(x2,y2),D(x0,y0),易知x1≠x2,x0y0則ax12+by12=1,ax22+by22=1,兩式作差有a(x1-x2)(x1+x又x1+x2=2x0,y1+y2=2y0,kOD=y0x0,∴ax0by0∴橢圓C的方程為ax2+2ay2=1,且a>0,即x21a設(shè)橢圓的長(zhǎng)半軸長(zhǎng)、短半軸長(zhǎng)分別為a1,b1,則a1故橢圓C的離心率e=113.34解析設(shè)直線(xiàn)x+y-1=0與橢圓x24+y23=1的交點(diǎn)坐標(biāo)為A(x1,y1),B(x2,y2則Mx1+x22,y1+y22因?yàn)锳,B在橢圓上,則x124+整理得y12-y22x12-x22=14.解(1)由題意ca=63,9a2+(2)設(shè)直線(xiàn)AB的方程為y=x+t,A(x1,y1),B(x2,y2),聯(lián)立x2+3y2=12,y=x+t,消去y得4x2+6tx+3t2-則x1+x2=-3t2,x1x2=3t2-124,則AB的中點(diǎn)M的橫坐標(biāo)為xM=-依題意知CM⊥AB,即kCM·kAB=-1,即yM-2xM+3×1=t4-2-3t4+3=-1,解得t=2,滿(mǎn)足Δ>∴|AB|=1+kAB2·(x1+x2)2-4x∴S△PAB=12|AB|·d=615.解(1)設(shè)A(x1,y1),B(x2,y2),則M(x1即kOM=y1+y2x1+x2=-12.因?yàn)锳,B在橢圓C上,又kAB=y1-y2x1-x2=1,所以1a又因?yàn)闄E圓C過(guò)點(diǎn)1,62,所以1a2+32b2=1,解得a2=所以橢圓C的標(biāo)準(zhǔn)方程為x24+(2)不存在.理由如下,由題意可知,直線(xiàn)l的方程為y=x+1.假設(shè)橢圓C上存在P,Q兩點(diǎn),使得P,Q關(guān)于直線(xiàn)l對(duì)稱(chēng),設(shè)P(x3,y3),Q(x4,y4),PQ的中點(diǎn)為N(x0,y0),所以