圓錐曲線設(shè)而不求法典型試題

圓錐曲線設(shè)而不求法典型試題例1,弧ADB為半圓，AB為直徑，0為半圓的圓心，且0D垂直于AB,Q為半徑0D的中點(diǎn)，已知AB長為4,曲線C過Q點(diǎn)，動點(diǎn)P在曲線C上運(yùn)動且始終保持/PA/+/PB/的值不變。過點(diǎn)D的直線與曲線C交于不同的兩點(diǎn)M、N，求三角形OMN面積的最大值。2例2：已知雙曲線x-y/2=1，過點(diǎn)M(1,1)作直線L,使L與已知雙曲線交于點(diǎn)，問：這樣的直線是否存在？若存在，求出L的方程；若不存在，說明原因。解：假定存在知足題意的直線L,設(shè)Qi(Xi,Yi),Q2(X2,Y2)2222代人已知雙曲線的方程，得xi-yi/2=1①，x2-y2/2=1②②-①，得(X-x)(x+x)-(y-yi)(y+y)/2=02i2i22io當(dāng)Xi=x2時，直線L的方程為x=i，此時L與雙曲線只有一個交點(diǎn)(i,0)不知足題意;當(dāng)xi孜2時，有(y2-yi)/(x2-xi)=2(x2+xi)/(y2+yi)=2.故直線L的方程為y-1=2(x-1)222-4X+3=0,其鑒別式查驗：由y-1=2(x-1),x-y/2=1，得2X"=-8<0,此時L與雙曲線無交點(diǎn)。綜上，不存在知足題意的直線3例3,已知，橢圓C以過點(diǎn)A(1,-),兩個焦點(diǎn)為(一1,0)(1,0)o2求橢圓C的方程；(2)E,F是橢圓C上的兩個動點(diǎn)，如果直線AE的斜率與AF的斜率互為相反數(shù)，證明直線EF的斜率為定值,2x2y解由題意，c=1,可設(shè)橢圓方程為-b24b2i1因為A在橢圓上，所以9,解得b2=3,b21b212x224b所以橢圓方程為一'1.43

3(舍去)。4(H)證明設(shè)直線AE方程：得yk(x1)3，代入7并求出這個定值。(3+4k22k)2)x+4k(32k)x4(12023設(shè)E(XE,YE),F(XF,YF)?因為點(diǎn)A(1,一)在橢圓上,2324(k)2123所以XE2yEkxEk。2又直線AF的斜率與2k代k，可得AE的斜率互為相反數(shù)，在上式中以34k4(2k)2123k。2XF34k2，YkxF2F所以直線EF的斜率kEFYYk(xFx)2k1FEEXXXX2FEFE即直線EF的斜率為定值,其值為1。2224,已知直線x2y20經(jīng)過橢圓C:XCS2a1(ab0)的左極點(diǎn)A和上極點(diǎn)，橢圓的右極點(diǎn)為，點(diǎn)DB和橢圓C上位于X軸上方的動點(diǎn)，直線,AS,BS與直線|:x10分別交于3求橢圓C的方程；求線段MN的長度的最小值；解方法一(1)由已知得，橢圓C的左極點(diǎn)為A(2,0),2上極點(diǎn)為D(0,1),a2,b1x2彳故橢圓C的方程為7y1(2)直線的斜率故可設(shè)直線的方程為yk(x2)，ASk顯然存在，且k0，AS進(jìn)而M(?曙yk(x2)x22“得(127y12224k)x16kx16k40，3316k24/曰鯊，進(jìn)而y14k設(shè)S(X,yJ,則(2),x廠得214k14k214k2S(28k企),又B(2,0)1即4k2'14k丄(x102)4k310丄)故|MN|16k1得10%13k33k33k016k1|MN|3k,3當(dāng)且僅當(dāng)型丄，即丄時等號建立433kk1時，線段MN的長度取最小值4例5?已知點(diǎn)A(X1,y1),B(X2,y2)(X1X20)是拋物線y2px(p0)上的兩個動點(diǎn)，0是坐標(biāo)原點(diǎn)，向量UUUUUIUuuuuuuuuuuuu22OAOBC的方程為Xy(X1212)y0OAOB知足OAOBX)X(yy,證明線段AB是圓C的直徑；(2)當(dāng)圓C的圓心到直線X-2Y=0的距離的最小值為時，求p的值￡3uuuuuuuuuuuuuuu25uuu2UUUUUU解析：(I)證明：QOAOBOAOB)(OAOB)21OB,(OAUUU2uuuuuu22UUUuuuuuumuUULOA2OAOBuuuuuuOA2OAOB2整理得：OAOBOBXXyy0OB120,12UULT，UUL設(shè)M(x,y)是以線段AB為直徑的圓上的隨意一點(diǎn)則MAT(xxj(xX)2即2整理得:X?(yyJ(yy)0MB220'1212)yy(XyX)x(y故線段AB是圓C的直徑UUUuuu2uuuuuuuuuuuuuuuuuu2證明2：QOAOBOAOB,(OAOB)(OA2UUUuuuuuuUUUUUUuuuuuuuu2OB)222OA2OAOBOBOA2OAOBOBuuuuuu0整理得：OAOBXxyi目20..(1)i設(shè)(x,y)是以線段AB為直徑的圓上則即31311(xXi,XX2)XX2xX1去分母得：(XX1)(xX2)(yy1)(yy?)0點(diǎn)(X,yd,y),(x,『1)(x,y)知足上方程，展開并將(1)代入得:1222222(X121y2)y0XyX)X(y故線段AB是圓C的直徑uuuuuuuuuuuuuuuuuuuuu證明3：QOAOBuuu22UUU2uuuuumUUU2OAOB,(OAOB)2(OAOB)22UUUUUUUUW2OAOA2OAOBOBuuuuuu整UUUOBOB2OA理得：OAOB0XXy1y20(1)12以線段AB為直徑的圓的方程為Xir)Xn2(yy1y2)2122(x2[(XX)(yy)]展開并將(1)代入得41212：x2y2(x-ix2)x(yiy2)yo故線段AB是圓C的直徑解法1:設(shè)圓C的圓心為C(X,y),則捲x22y1y222px1,y22Qy1222px2(p0)y2X!X24p2Xy1y2又因2X1X2y1y222yiy224pQXx20,yiy2|yiy24p2Xii22iy22%y2)學(xué)x(yiy2護(hù)2214p4pi22(y22p2)c2ppx2p設(shè)圓心C到直線x-2y=0的距離為d,則所以圓心的軌跡方程為i22dlx2y|d.l5—(y22p2)2y|y22py2p2|pl(yP)2p2|75p25當(dāng)y=p時,d有最小值-匕,由題設(shè)得、5V5V5p2.解法2:設(shè)圓C的圓心為C(x,y),則xix22yiy?22Qyi20)2px,y2px(pi22x|x222yiy2又因xX24p2yiy20xx2yiy2|22yiy2Qx|x24p20,yiy2yiy24p2x亠i,22"2(y4p2iy2i4p22-(y22p2)p所以圓心的軌跡方程為設(shè)直線px2p2x-2y+m=0到直線x-2y=0的距離為m2因為x-2y+2=0與y2px2p2無公共點(diǎn)所以當(dāng)x-2y-2=0與y2px2p2僅有一個公共點(diǎn)時，該點(diǎn)到直線x-2y=0的距離最小值為亠55x2y20L(2)y2px2p2L⑶將⑵代入⑶得y22py2p22p0224p4(2p2p)0Qp0p2.解法3:設(shè)圓C的圓心為C(x,y),則yiy2圓心C到直線x-2y=0的距離為d,則XiX2(yy2)I2i22Qyi2pX』22px2(p0)i22