①若實(shí)數(shù)a,b分別滿足f(a)=0,f(b)=0,由此a,b可視為方程f(x)=0的兩個(gè)根.--雙切線、斜率和(積)為②如果A(x1,y1),B(x2,y2)滿足的方程結(jié)構(gòu)相同，則A,B為方程所表示的--切點(diǎn)弦方程推導(dǎo)的核心思路.(1)+y2=1(2)設(shè)點(diǎn)A(x1,y1(，B(x,y(，C(x3,y3(，D(x4,y4(，2y1-x1y2=y2-y1，直線CD的方程為y--=x-，(2x1-3((y4-y3(令y=0得，x=-y1(x4-x3(+((2x1-3((y4-y3(=-y1-+(3x1-4(-(2x1-3(-==-y1(3x2-4(+y2(3x1-4(4(y1-y2(+3(x1y==2x1-3(-y1(2x2-3(3(y1-y2(+2(x1y2-x2y1(5，(3)求證直線過定點(diǎn)(x0,y0(，常利用直線的點(diǎn)斜式方程y-y0=k(x-x0(或截距式y(tǒng)=kx+b來證明.上，圓E:(x-2)2+y2=r2(其中0<r<1).(1)若r=,Q為圓E上的動(dòng)點(diǎn)，求線段PQ長(zhǎng)度的(2)設(shè)D(1,t(是拋物線C1上位于第一象限的一點(diǎn)，過D作圓E的兩條切N.證明：直線MN經(jīng)過定點(diǎn).(2)根據(jù)兩點(diǎn)坐標(biāo)可得直線MN,DM的直線方程，由直線與圓相切可得a,b是方程(r2-1(x2+(2r2-4(x+設(shè)點(diǎn)P(t2,t(，則|PQ|≥|PE|-=(t2-2(2+t2-=(t2-2+-≥,(2)∵D(1,t(是拋物線C1上位于第一象限的點(diǎn)，設(shè)M(a2,a(,N(b2,b(，則：直線MN:y-a=(x-a2(，即y-a=(x-a2(，即x-(a+b(y+ab=0.直線DM:y-1=(x-1(，即x-(a+1(y+a=0.由直線DM與圓相切得=r，即(r2-1(a2+(2r2-4(a+(2r2-4(=0.同理，由直線DN與圓相切得(r2-1(b2+(2r2-4(b+(2r2-4(=0.所以a,b是方程(r2-1(x2+(2r2-4(x+(2r2-4(=0的兩個(gè)解，代入方程x-(a+b(y+ab=0得(x+2y+2(r2+(-x-4y-4(=0，∴直線MN恒過定點(diǎn)(0,-1(.找到定點(diǎn).技巧：若直線方程為y-y0=k(x-x0(,則直線過定點(diǎn)(x0,y0(;若直線方程為y=kx+b(b為定值),則直(3)過點(diǎn)A的且斜率存在的直線l1,l2分別與橢圓交于點(diǎn)P,Q(均異于點(diǎn)A)，若點(diǎn)B到直線l1,l2的距離相(3)直線PQ過定點(diǎn)(-6,-3(，證明見解析可得c2=a2-b2=3，又A(-2,1(為橢圓上一點(diǎn)，43+b2643+b262所以橢圓E的方程為+22y32y313-2<k<13-2<k<-2-3此時(shí)直線l的方程為y-1=-(x+2(，即x+y+1=0；直線l1:y-1=k1(x+2(,l2:y-1=k2(x+2(，(1+2k(x2+(8k+4k1(x+8k+8k1-4=0，可得直線PQ的方程為化簡(jiǎn)得(2-k1+2k(y-9k1=(k1+1+k(x，所以(2y-x((1+k(+k1(-y-x-9(=0，由解得，可得直線PQ過定點(diǎn)(-6,-3(.4.如圖，在平面直角坐標(biāo)系xOy中，雙曲線=1(a>0,b>0(的上下焦點(diǎn)分別為F1(0,c(，F(xiàn)(2)設(shè)A,B是雙曲線上位于y軸右方的-x2=1(1)將點(diǎn)的坐標(biāo)代入雙曲線的方程求解即可；將點(diǎn)(0,和代入雙曲線方程得：雙曲線的方程為-x2=1.(2)(i)設(shè)A(x1,y1(,B(x2,y2(,B關(guān)于原點(diǎn)對(duì)稱點(diǎn)記為C(x3,y3(，則x3=-x2,y3=-y2.又因?yàn)锳F1kFA=kBFFA=kCF，故A,F1,C三點(diǎn)共線.-|BF2|=|AF1|-|CF1|=2.直線AF1與雙曲線上支有兩個(gè)交點(diǎn)，所以Δ>0,x1x3=解得|k|<.4+7k2-4=0，且由圖可知k>0，即(2k2-1((k2+4(=0， =2=22+=22=22++ + 12k2-4\(k2-2(2k2-2 |k2-2|1\(k2-2(2k2-2 |k2-2|111+k21++得解.25.已知雙曲線C：x2-y2=1(a>0,b>0(的離心率為2，點(diǎn)(3,-1(在雙曲線C上．過C的左焦點(diǎn)2(3)點(diǎn)P(-4,2(，直線AP交直線x=-2值.x2-x2-y22=x2-y22=1.(2)雙曲線C的左焦點(diǎn)為F(-4,0(，-1(y2-8my+8=0，顯然m2-1≠0，Δ=64m2-32(m2-1)=32(m2+1)>0，設(shè)A(x1,y1(，B(x2,y2(，則y1+y2=,y1y2=<0，得-1<m<1，于是M=(x1+2,y1(,M=(x2+2,y2(，M?M=(x2+2((x1+2(+y1y2=(my1-2((my2-2(+y1y2即MA?MB≠0，因此MA與MB不垂直，所以不存在直線l，使得點(diǎn)M在以AB為直徑的圓上.(3)由直線AP:y-2=k1(x+4(，得Q(-2,2+2k1)，=-2my2-2y1+4+2my1+2mk1y1my1(my2-2(,1my1=y1-2，且y1+y2=my1y2，①求證：∠AQP=∠BQP；2=4x2=2px(p>0),由a2-b2=4-3=1，得c=1.由此能求出拋物線D的方(2)①設(shè)A(x1,y1(，B(x,y(，由于O為PQ中點(diǎn)，則Q(-4,0(,故當(dāng)l⊥x軸時(shí)由拋物線的對(duì)稱性知∠AQP=∠AQP=∠BQP. ∴拋物線D的方程為y2=4x；(2)①設(shè)A(x1,y1(，B(x,y(，由于O為PQ中點(diǎn)且P(4,0(，則Q(-4,0(，當(dāng)l不垂直x軸時(shí)，顯然直線l的斜率不為0，設(shè)l:y=k(x-4((k≠0(，x1+x2x1+4x2+4(x1+4((x2+4(，所以kAQ+kBQ=k(x1-4(+k(x2x1+4x2+4(x1+4((x2+4(，則∠AQP=∠BQP，②設(shè)存在直線m:x=t滿足題意，設(shè)圓心M,，過M作直線x=t的垂線，垂足為E，圓M與直線m的一個(gè)交點(diǎn)為G，2-|ME|2，即|EG|2=|MA|2-|ME|2=-t(2=y++t(x1+4(-t2=x1-4x1+t(x1+4)-t2=(t-3(x1+4t-t2，此時(shí)直線m被以AP為直徑的圓截得的弦長(zhǎng)恒為定值2(5)代入韋達(dá)定理求解.于M，直線PB交y軸于N.(1)(-∞,-3(∪(-3,0(∪(0,1((2)-1若直線l與拋物線的一個(gè)交點(diǎn)為(1,-2)，此時(shí)該點(diǎn)與點(diǎn)P所在的直線此時(shí)=-3，所以直線l斜率k≠-3.故直線l的斜率k的取值范圍是k∈(-∞,1)且k≠-3且k≠0.即率k的取值范圍是(-∞,-3(∪(-3,0(∪(0,1(.則y1+y2=4m,y1y2=-4，∵D=λA，D=μB，∴λ=-1-，μ=-1-，∴λ+μ+=-2-∴λ+μ=-1.m-1=-λ,故λ=1-ym，由Q=μQ得μ=1-yn，設(shè)A(x1,y1(，B(x,y(，直線PA方程為y-2=m=由直線PB可得yn=，(5)代入韋達(dá)定理求解.F(2)過點(diǎn)P(0,2、3(的直線l交C于M、N兩點(diǎn)(M位于P與N之間)，記△PMB1、△PNB1的面積分別為(1)+=1(2),設(shè)M(x1,y1(，N(x,y(，3μ|x2|(5)代入韋達(dá)定理求解.AB的距離為|OB|(O為坐標(biāo)原點(diǎn)).(2)若橢圓則稱橢圓E為橢圓C的λ倍相似橢圓．已知橢圓E是橢圓C(2)證明見解析.(1)∵A(-a,0(,B(0,b(，∴F1(-1,0)到直線AB的距離為d==b，2=7(a-1)2，2=a2-12)3)4)，(8km)2-4(3+4k2)(4m2-12)=48(4k2+3-m2)>0，(*)∴|x1-x2|=(x1+x2)2-4x1x2=43(4k2+32-m2)將y=kx+m代入橢圓E的方程得(3+4k2)x2+8kmx+4m2-36=0，x3-x4|=由MQ+PQ=2NQ可得NM=PN，∴|MQ|=3|PN|，所以|x3-x4|=3|x1-x2|，2-42=1(1)+=1R代入x=r到+=1，不妨設(shè)C(r,、16-r2(，設(shè)E為AC與圓O的切則由切線的性質(zhì)，CE=CF=16-r2，OE=OF=r，故AE=AO2-OE2=16-r2，故AC=AE+(3)設(shè)圓O的切線方程為y-2=k(x-22(，即kx-y+2-22k=0.聯(lián)立(x-2(，則(1+4k2(x2+82k(1-2k(x+8(4k2-4k-1(=0.設(shè)Q(x1,y1(,R(x2,y2(，則22x1=8(4k-4k1-1(，即x1=22(4k-4k1-1(=-8-2x2=1-y2=k1(x1-22(-k2(x2-22(=k1x1-k2x2-22(k1-k2(=.又kQR=故直線QR的方程為y-y1=(x-x1(，即(y1-y2(x-(x1-x2(y+x1y2-x2y1=0，故O到直線QR的距離d=|x1y2-x2y1|=(y1-y2(2+(x1-x2(2 |x1[k2(x2-22(+2[-x2[k1(x1-22(+2[|(y1-y2(2+(x1-x2(2= |(k2-k1(x1x2+2、2(x2k1-x1k2(+、2(x1-x2(|(y1-y2(2+(x1-x2(2=2|PF|.(2)過曲線C上的點(diǎn)M(x0,y0((x0≥1(作圓(x+1(2+y2=1的兩條切線，切線與y軸交于A，B，求△MAB面積的取值范圍.(2)設(shè)切線方程y-y0=k(x-x0(，通過點(diǎn)到切線的距離，化簡(jiǎn)成k|PQ|=2|PF|，得|x+4|=2、(x+1(2+y2，兩邊平方得+=1，2+=1；(2)設(shè)點(diǎn)M(x0,y0(的切線方程為y-y0=k(x-x0((斜率必存在)，圓心為F(-1,0(，r=1所以F(-1,0(到y(tǒng)-y0=k(x-x0(的距離為：d=|-k+y0-kx0|=11+k2平方化為(x+2x0(k2-2(x0+1(y0k+y-1=0，設(shè)PA，PB的斜率分別為k1，k2x+2x0，12x+2x0則k1+k2=2(x0+x+2x0，12x+2x0因?yàn)镻A：y-y0=k1(x-x0(，令x=0有yA=y0-k1x0，同理yB=y0-k2x0所以|AB|=|yA-yB|=x0|k1-k2|=x0(k1+k2(2-4k1k2=又因?yàn)?y=12-3x代入上式化簡(jiǎn)為|AB|=令f(x(=，x∈[1,2[，求導(dǎo)知f(x(在x∈[1,2[為增函數(shù)，所以S∈,2.x-1)2+(y+1)2=和拋物線C2:x2=4y,P(x0,y0(是圓C1上一點(diǎn)，M是拋物線C2上一2(1)M,或M,；(2)證明見解析.(1)焦點(diǎn)F坐標(biāo)為，設(shè)M(xM,利用圓的切線長(zhǎng)公式、拋物線的定義建立方程求解即得；線AB方程并化簡(jiǎn)整理為x-y0，利用已知面積得到x-4y0=3，與(x0-1(2+(y0+1(2=聯(lián)立得(x0-1((x+x+19x0-13(=0，然后利用零點(diǎn)存在定理判定解的個(gè)數(shù)即可.(1)焦點(diǎn)F坐標(biāo)為，設(shè)M(xM,(，則|PM|=所以xM=或xM=，所以M,或M,，(2)設(shè)P(x0,y0(，則(x0-1(2+(y0+1(2=，設(shè)直線PA方程為y-y0=k1(x-x0(，代入x2=4y，得x2-4k1x-4(y0-k1x0(=0,Δ=16k+16(y0-k1x0(=0，整理得k-k1x0+y0=0①,同理，直線PB方程為y-y0=k2(x-x0(，有k-k2x0+y0=0②,,k2是方程k2-kx0+y0=0的兩根，2-4k1x-4(y0-k1x0(=0中，xA+xA=4k1，則xA=2k1所以A(2k1,k(，同理B(2k2,(,kAB=直線AB方程為y-k=(x-2k1(即y=x-k1k2即y=x-y0|AB|=1+|2k1-2k2|=、4+x?(k1+k2(2-4k1k2=、4+x?、x-4y0所以x-4y0=3，與(x0-1(2+(y0+1(2=聯(lián)立得(x0-1((x+x+19x0-13(=0長(zhǎng)和拋物線的定義建立方程求解是第一問中的關(guān)鍵；第二問中關(guān)鍵點(diǎn)由同構(gòu)方程k-k1x0+y0=0，k-k2x0+y0=0，知k1,k2是方程k2-kx0+y0=0的兩根，從而得到k1+k2=x0,k1k2=y0;利用零點(diǎn)存在定理判定三次函數(shù)在給定區(qū)間上的零點(diǎn)個(gè)數(shù)問題.【解析】(1)將y=代入C:x2=2py(p>0(中得x=±、p，2=2yy0=x0-2，設(shè)M(x1,y1(,N(x2,y2(，則直線GM,GN的方程分別為y-y1=x1(x-x1(,y-y2=x2(x-x2(，而直線MN的方程為y-y1=(x-x1(=(x-x1(，即+y1=x0x-y0-+y1=x0x-y0=x0x-x0+2=x0(x-1(+2，則直線MN過定點(diǎn)(1,2)；設(shè)直線PQ:y=kx+(k≠0(,P(x3,y3(,Q(x4,y4(，聯(lián)立得，得x2-2kx-1=0，則x3+x4=2k,x3x4=-1，聯(lián)立{，解得H(k,-，故kFH?kPQ=-1，即PQ⊥FH，由P=λF，得x3=-λx4，結(jié)合根與系數(shù)的關(guān)系可知x4=-,-2=-1，△HPQ=|PQ|?|HF|=(1+k2，由于-2(在λ∈時(shí)為增函數(shù)，|PD||PE|=|AD||BE|.(1)x2=4y(1)設(shè)A(x1,y1(，B(x,y((x1<0<x2(，直線方程聯(lián)立拋物線方程，利用韋達(dá)定理表示y1+y2,y1y2，結(jié)合算即可證明.(1)設(shè)A(x1,y1(，B(x,y(，(x1<0<x2(，2I=PB=，則直線PA方程為y-y1=(x-x1)，即x1x=2(y+y1(，同理直線PB方程為x2x=2(y+y2(.-x2(=2(y1-y2(，兩式相加得2y=(x1+x2(-(y1+y2(=(y1-1+y2-1(-(y1+y2(=-2，即y=-1，所以點(diǎn)P(2,-1(.設(shè)直線DE與拋物線相切于點(diǎn)T(x0,y0(，則直線DE方程為xx0=2(y+y0(.因?yàn)閨PD||PE|=1+|yD-yP|?1+|yE-yP|=1+1+(yD+1((yE+1(，故要證|PD||PE|=|AD||BE|，即證yDyE+yD+yE+1=y1y2-y2yD-y1yE+yDyE，即證yD+yE+yDy2+y1yE=0，即證4x0(x1+x2(+x0x1x2(x1+x2(=0，4+x1x2((x1+x2(=0，1x2<0故|PD||PE|=|AD||BE|.:y=kx+m(k≠±2(與C有唯一的公共點(diǎn)M，過M且與l1垂直的直線分別交x軸，y軸于點(diǎn)A(x,0(,B(0,y(兩點(diǎn)，當(dāng)M運(yùn)動(dòng)時(shí)，求點(diǎn)D(x,y(的軌跡方程；唯一的公共點(diǎn)M，過M且與l1垂直的直線分別交x軸，y軸于點(diǎn)A(x,0(,B(0,y(兩點(diǎn)，即可求解；(1)聯(lián)立方程，得(4-k2(x2-2kmx-m2-16=0(k≠±2(，與C有唯一的公共點(diǎn)M，所以Δ=(-2km(2-4(4-k2((-m2-16(=0，過M且與l1垂直的直線為(，則x=-20×,y=-，即-=1(y≠0(，所以D的軌跡方程為=1(y≠0(.2:x=ty-、3，3ty-1=0,Δ=16t2+16>0，23t-13ty-1=0,Δ=16t2+16>0，23t-1t2+4,y1y2=t2+4，設(shè)P(x1,y1(,Q(xt2+4,y1y2=t2+4，|PQ|=1+t2|y1-y2|=1+t2(y1+y2(2-4y1y2(t2+4(2=1+t2+4=4(t2+(t2+4(2=1+t2-x1-x2橢圓在x軸上方對(duì)應(yīng)方程為y=1-2,y-x1-x244x1則點(diǎn)P處切線斜率為=-x4-x14x1則點(diǎn)P處切線斜率為=-x4x1x4(x-x1(，即x1x4(x-x1(，即+y1y=1，4同理可得點(diǎn)Q處的切線方程為x2x+y2y=1，4x1x4x2x4由(4(y2-y1(x1x4x2x4由(4(y2-y1(x1y2-x2y1+y1y=1①,+y2y=1②,得xN===-(ty1-3(y2-(ty2-3((ty1-、3(ty1-y11+3、3(ty1-y11+3x11+=3代入①得yN==y133,-t(，而P=(x2-x1,y2-y1(=((ty2-3(-(ty1-3(,y2-y1(=(y2-y1((t,1(，=?|PQ|?|NF1|=??=?.所以f(m(在[1,+∞(上單調(diào)遞增，三角形面積最值.(2)設(shè)曲線O:x2+y2=1(x≠0(的切線l與橢圓C交于A,B兩點(diǎn)，且以A,B為切點(diǎn)的橢圓C的切線交于0=±2及x0≠0,x0≠±2，根據(jù)直線與橢圓的位置關(guān)系結(jié)合判別式計(jì)算即可證明；合(1)得過A,B的橢圓切線方程，聯(lián)立兩切線方程求交點(diǎn)得M坐標(biāo)與A,B,P坐標(biāo)的關(guān)系，再結(jié)合點(diǎn)在圓上消元化簡(jiǎn)得xM=4x0,yM=y0，根據(jù)三角形面積S△ABM=PM|y1-y2|結(jié)合導(dǎo)數(shù)求其值域即可；法二、設(shè)AB:x=利用弦長(zhǎng)公式及點(diǎn)到直線的距離公式計(jì)算面積求范圍即可.0=0x2--1=0，Δ=-4+-1(==0，∴x+y0y=1與橢圓C只綜上，x+y0y=1是橢圓C在P(x0,y0(處的切線方程；若x0≠±1，可設(shè)l:y-y0=k(x-x0(，由直線與圓的位置關(guān)系知：kOP?k=-1?k=-，則l:y0y-y+x0x=x?x0x+y0y=1，0x+y0y=1，若x0=-1，顯然切線方程為x=-1，滿足x0x+y0y=1，故圓在切點(diǎn)P處的切線方程為x0x+y0y=1；0≠0,x0≠±1，設(shè)A(x1,y1(,B(x2,y2(,M(xM,yM(，x0x01-y2=y0(x2-x1(，x+y2y=x+y2y=1，x14x+y1y=1x2x+x14x+y1y=1x2x+(y1-y2(y=0，x+y2y=144∵x0≠0,x0≠±1,∴x1-x2≠0，得yy0,yM=x1y0,yM=x1x0+y1y0因?yàn)锳(x1,y1(，點(diǎn)在切線x0x+y0y=1上，所以x1x0+y1y0=1，得xM=4x0,yM=使用水平底鉛垂高計(jì)算△ABM的面積，鉛垂高為|y1-y2|=又(x1-x2(2=(x1+x2(2-4x1x2=即|y1-y2|= x3x+1 3x x3x+1 3x+1S△ABM=x0∵點(diǎn)P在圓O上，∴x0∈[-1,1[，由題意，x0≠0，(0<x<1(,fI(x(=3x2(2(0<x<1(,fI(x(=3x2(2+12(△ABM∈0,>0，f(0(=0,f(1(=33(3x+1(,代入橢圓C的方程得y=±y1-y2|=,代入橢圓C的方程得y=±y1-y2|=3,同理x0,同理x0=-1時(shí)，S△ABM=33.(s.綜上S△ABM(s.法二、設(shè)直線AB:x=ty+m，t2+12=t2+1≥1.設(shè)A(x1,y1),B(x2,y2),M(x0,t2+1(t2+4)y2+2tmy+m2-4=0，t2+443?1+t2=t2+4.Δ=4t2m2-4(t2+4)(m2-4)=16(t2-m2+4)則|AB|=1+t2?(t2+443?1+t2=t2+4.x1x+y1y=1過點(diǎn)A的橢圓切線為x1x+y1y=1，過點(diǎn)x1x+y1y=1(x+y2y=1，4所以A，B的坐標(biāo)滿足直線x0x+y0y=4-則點(diǎn)P到直線AB的距離d=|+-m|=3，1+t2|m|1+t2所以S△MAB=|AB|d=|m|6(t24)=|m|+3)，ΔMAB→0，1.(2024·遼寧·模擬預(yù)測(cè))已知雙曲線C:-=1(a>0,b>0(過點(diǎn)M(2,3(，離心率為2.MT的斜率之積為定值.2-=1【解析】(1)由雙曲線C:-=1(a>0，b>0)過點(diǎn)M(2,3(，則-=1，2=12=3，2-則M(2,3(平移到M1(0,0(，Q(-4,6(平移到Q1(-6,3(，S1(x1,y1(2-y2+(12x-6y((mx+ny(=0，則(6n+1(y2-(12n-6m(xy-(3+12m(x2=0，2-(12n-6m(-(3+12m(=0，則=kM1S1?kM1又直線S1T1過Q1(-6,3(，則-6m+3n=1，即6m=3n-1，2；2.(2024·云南·模擬預(yù)測(cè))拋物線Γ:y2=2px(p>0(的圖象經(jīng)過點(diǎn)M(1,-2(，焦點(diǎn)為F，過點(diǎn)F且傾斜角為(1)y2=4x(2)|AB|=(1)曲線y2=2px圖象經(jīng)過點(diǎn)M(1,-2(，所以(-2(2=2p，所以p=2，由|AB|=x1+x2+p=所以弦|AB|=.A(x1,y1(，B(x2,y2(，C(x3,y3(，D(x4,y4(，y2=-4.因此直線CD的方程為x=2m(y-y3(+x(3)求證直線過定點(diǎn)(x0,y0(，常利用直線的點(diǎn)斜式方程y-y0=k(x-x0(或截距式y(tǒng)=kx+b來證明.(2)設(shè)A,B是雙曲線上位于y軸右方的-x2=1(1)將點(diǎn)的坐標(biāo)代入雙曲線的方程求解即可；將點(diǎn)(0,(和(e,代入雙曲線方程得：雙曲線的方程為-x2=1.(2)(i)設(shè)A(x1,y1(,B(x2,y2(,B關(guān)于原點(diǎn)對(duì)稱點(diǎn)記為C(x3,y3(，則x3=-x2,y3=-y2.又因?yàn)锳F1kFA=kBFFA=kCF，故A,F1,C三點(diǎn)共線.-|BF2|=|AF1|-|CF1|=2.直線AF1與雙曲線上支有兩個(gè)交點(diǎn)，所以Δ>0,x1x3=解得|k|<.|-|C|x1|-|x3|=(x1+x3(=，4+7k2-4=0，且由圖可知k>0，即(2k2-1((k2+4(=0， 11+=+==(k2-k2-211++得解.4.(2024·福建南平·模擬預(yù)測(cè))已知拋物線C:y2=2px(p>0(的準(zhǔn)線l與圓O:x2+y2=1相切.積的最小值.2=4x(2)設(shè)直線PA、PB的方程分別為y-y0=k1(x-x0(、y-y0=k2(x-x0(，可得S△PAB=(x1+1(2.(k1+k2(2-4k1k2(x0>1(，過點(diǎn)P的圓O的切線方程與圓相切可得(x-1(k2-2x0y0k+y-1=-所以C的方程為y2=4x；(2)由(1)知準(zhǔn)線l的方程為x=-1，設(shè)直線PA的方程為y-y0=k1(x-x0(，當(dāng)x=-1時(shí)，yA=k1(-1-x0(+y0，設(shè)直線PB的方程為y-y0=k2(x-x0(，當(dāng)x=-1時(shí)，yB=k2(-1-x0(+y0，由題意得S△PAB=|yA-yB|?(x0+1(=|k1-k2|?(x0+1(2=(x0+1(2.、(k1+k2(2-4k1k2(x0>1(，設(shè)過點(diǎn)P(x0,y0(的圓O的切線方程為y-y0=k(x-x0(,則|-x0k+y0|=1，化簡(jiǎn)得(x-1(k2-2x0y0k+y-1=0，k2+1Δ=4(x+y-1(=4(x+4x0-1(=4[(x0+2(2-5[>0(x0>1(，=-.x+y-1，令x0-1=t(t>0(，所以S△PAB≥、(4+6(×(4+4(=4、5，為P.表示斜率后求解即可得.【解析】(1)由M:+y2=1可得A(-2,0(,C(0,1(，則kAC=tan∠PAB=tan2∠CAB=2×=4，23故lAP:y=(x+2(，令(x+2(=，解得x=-2或x=-，當(dāng)x=-2時(shí)，交點(diǎn)為點(diǎn)A，舍去，當(dāng)x=-時(shí)，y=×(-+2(=，即P(-(，則kPC==-kAP，故∠PAB=∠PBA，PA=PB，則x=-為∠APB的角平分線，聯(lián)立lAC:y=x+1，有y=×(-+1=，則點(diǎn)(-,為△PAB內(nèi)心，+1(x+8kx=0，則xQ=-，則yQ=k?(-+1=，即Q(-,，+(2k+1(x=0，則xP=-=1-，則yP=k?(-+1=-k+，即P(1-,k-，則kAP=-=-k，故lAP:y=-k(x+2(，y=-k(x+2(4+y=12+1y=-k(x+2(4+y=1則由x