圆锥曲线-答案

deserts__xuan
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  1. normal
    a2+b2−16=0  then  a2+b2=16a^2+b^2-16=0~~then~~a^2+b^2=16

弦长问题

  1. normal

    let  AB:y=x+3then  Δ=4⋅4⋅1⋅1(5−3)=32then  ∣AB∣=2325=85let~~AB:y=x+\sqrt3 \\then~~\Delta=4\cdot4\cdot1\cdot1(5-3)=32 \\then~~|AB|=\sqrt2\frac{\sqrt{32}}{5}=\frac{8}{5}

  2. normal

    let  l:x=ty+m  A(x1,y1)  B(x2,y2)∵ tangency∴ m2=1+t2then  Δ=4⋅4⋅1⋅1(4+t2−m2)=48then  ∣AB∣=1+t2484+t2=43⋅1+t2(1+t2)2+3≤2let~~l:x=ty+m~~A(x_1,y_1)~~B(x_2,y_2) \\\because~tangency \\\therefore~m^2=1+t^2 \\then~~\Delta=4\cdot4\cdot1\cdot1(4+t^2-m^2)=48 \\then~~|AB|=\sqrt{1+t^2}\frac{\sqrt{48}}{4+t^2}=4\sqrt3\cdot\frac{\sqrt{1+t^2}}{(\sqrt{1+t^2})^2+3}\le2

面积问题

  1. normal

    S=12d⋅PQy=kx−2→d=21+k2PQ=1+k2Δ4k2+1Δ=4⋅4⋅1⋅1(4k2+1−4)S=4⋅4k2−34k2+1=4⋅4k2−3(4k2−3)2+4=414k2−3+44k2−3≤1S=\frac{1}{2}d\cdot PQ\\ y=kx-2\to d=\frac{2}{\sqrt{1+k^2}}\\ PQ=\sqrt{1+k^2}\frac{\sqrt{\Delta}}{4k^2+1}\quad \Delta=4\cdot4\cdot1\cdot1(4k^2+1-4) \\S=4\cdot\frac{\sqrt{4k^2-3}}{4k^2+1}=4\cdot\frac{\sqrt{4k^2-3}}{(\sqrt{4k^2-3})^2+4}=4\frac{1}{\sqrt{4k^2-3}+\frac{4}{\sqrt{4k^2-3}}} \le1

  2. hard

∣AB∣=1+k2Δ12k2+1∣CD∣=1+k2Δ22k2+1→Δ1=Δ24⋅2⋅1⋅1((2k2+1)−m12)=4⋅2⋅1⋅1((2k2+1)−m22)→m1=−m2let  m1=m,  then  m2=−mh=2∣m∣1+k2→S=2∣m∣Δ2k2+1=42∣m∣2k2+1−m22k2+1≤42m2+2k2+1−m222k2+1=22|AB|=\sqrt{1+k^2}\frac{\sqrt{\Delta_1}}{2k^2+1}\\ |CD|=\sqrt{1+k^2}\frac{\sqrt{\Delta_2}}{2k^2+1}\\ \to \Delta_1=\Delta_2\quad \\ 4\cdot2\cdot1\cdot1((2k^2+1)-m_1^2)=4\cdot2\cdot1\cdot1((2k^2+1)-m_2^2)\to m_1=-m_2 \\let~~m_1=m,~~then~~m_2=-m \\h=\frac{2|m|}{\sqrt{1+k^2}}\to \\S=2|m|\frac{\sqrt{\Delta}}{2k^2+1}=4\sqrt2\frac{|m|\sqrt{2k^2+1-m^2}}{2k^2+1}\le4\sqrt2\frac{\frac{m^2+2k^2+1-m^2}{2}}{2k^2+1}=2\sqrt2
  • 面积的两种表示方式，一个是分割法，一个是直接底乘以高。
  • 一次比二次的函数求最值。

平面向量综合

这里不妨设两个交点A(x1,y1),B(x2,y2)A(x_1,y_1),B(x_2,y_2)

  1. normal

    let  AB:y=kx+2Δ>0  that  means  4k2+1−4>0→k2>34x1x2+y1y2=(k2+1)x1x2+2k(x1+x2)+4>0x1+x2=−2⋅k⋅4⋅24k2+1x1x2=4(4−1)4k2+1→k2<4overall  k∈(−2,−32)∪(32,2)let~~AB:y=kx+2 \\\Delta>0~~that~~means~~4k^2+1-4>0\to k^2>\frac{3}{4}%这是特别容易被忽视的一个细节 \\x_1x_2+y_1y_2=(k^2+1)x_1x_2+2k(x_1+x_2)+4>0 \\x_1+x_2=\frac{-2\cdot k\cdot 4 \cdot2}{4k^2+1}\quad x_1x_2=\frac{4(4-1)}{4k^2+1}\to k^2<4 \\overall~~k\in(-2,-\frac{\sqrt3}{2})\cup(\frac{\sqrt3}{2},2)

  2. noraml

    GA→=(my1+54,y1)GB→=(my2+54,y2)GA→⋅GB→=(m2+1)y1y2+54m(y1+y2)+2516y1+y2=4m4+2m2y1+y2=−64+2m2then  GA→⋅GB→=178m2+144+2m2>0■\overrightarrow{GA}=(my_1+\frac{5}{4},y_1)\quad \overrightarrow{GB}=(my_2+\frac{5}{4},y_2)\\ \overrightarrow{GA}\cdot\overrightarrow{GB}=(m^2+1)y_1y_2+\frac{5}{4}m(y_1+y_2)+\frac{25}{16} \\y_1+y_2=\frac{4m}{4+2m^2}\quad y_1+y_2=\frac{-6}{4+2m^2} \\then~~\overrightarrow{GA}\cdot\overrightarrow{GB}=\frac{\frac{17}{8}m^2+\frac{1}{4}}{4+2m^2}>0\quad\blacksquare

  3. 点评一下：这里面第一步难的是转化成∠MBP\angle MBP为锐角来做，这是第一个难点或者说是套路，第二个套路就是这里面设点的坐标设的是MM，虽然是有两个参量，如果设PP点的话，这道题基本就无了，两个套路点，好好把握。

    let  M(x0,y0)then  AM:yx+2=y0x0+2let  xp=4then  yp=6y0x0+2then  BM→⋅BP→=(x0−2,y0)(2,6y0x0+2)=6y02+2x02−8x0+2we  note  that:y02=3(1−x024)then  BM→⋅BP→=52(2−x0)>0■let~~M(x_0,y_0)\quad then~~AM:\frac{y}{x+2}=\frac{y_0}{x_0+2} \\let~~x_p=4\quad then~~y_p=\frac{6y_0}{x_0+2} \\then~~\overrightarrow{BM}\cdot\overrightarrow{BP}=(x_0-2,y_0)(2,\frac{6y_0}{x_0+2})=\frac{6y_0^2+2x_0^2-8}{x_0+2} \\we~~note~~that:y_0^2=3(1-\frac{x_0^2}{4}) \\ then~~\overrightarrow{BM}\cdot\overrightarrow{BP}=\frac{5}{2}(2-x_0) >0\quad\blacksquare

  4. 常规题

    let  M(x1,y1)  N(x2,y2)then  MB:y+2x=y1+2x1let yG=1then xG=3x1y1+2we  note  that  AGN  collinear∴kAG=kANin  other  words−13x1y1+2=y2−2x2in  addition:y=kx+4Bring  in,simplify:4kx1x2+6(x1+x2)=0x1+x2=−64k8k2+4x1x2=968k2+4Bring  in  valid■let~~M(x_1,y_1)~~N(x_2,y_2) \\then~~MB:\frac{y+2}{x}=\frac{y_1+2}{x_1} \\let~y_G=1\quad then~x_G=\frac{3x_1}{y_1+2} \\we~~note~~that~~AGN~~collinear \\\therefore k_{AG}=k_{AN}\quad in~~other~~words\quad \frac{-1}{\frac{3x_1}{y_1+2}}=\frac{y_2-2}{x_2} \\in~~addition:y=kx+4 \\Bring~~in,simplify:4kx_1x_2+6(x_1+x_2)=0 \\x_1+x_2=\frac{-64k}{8k^2+4}\quad x_1x_2=\frac{96}{8k^2+4} \\Bring~~in~~valid\quad\blacksquare

动直线定点

点评：套路就是一定要去设这条直线的方程，而不要去设其它的方程

  1. normal

    let  A(x1,y1)  B(x2,y2)AB:y=kx+bwe  note  that  kCA+kCB=kx1+b−1x1+kx2+b−1x2=−1simplify  (2k+1)x1x2+(b−1)(x1+x2)=0x1+x2=−8kb4k2+1x1x2=4(b2−1)4k2+1then  2kb−2k+b2−1=(2k+b+1)(b−1)=0→2k+b+1=0overall  (2,−1)■let~~A(x_1,y_1)~~B(x_2,y_2)\quad AB:y=kx+b \\we~~note~~that~~k_{CA}+k_{CB}=\frac{kx_1+b-1}{x_1}+\frac{kx_2+b-1}{x_2}=-1 \\simplify~~(2k+1)x_1x_2+(b-1)(x_1+x_2)=0 \\x_1+x_2=\frac{-8kb}{4k^2+1}\quad x_1x_2=\frac{4(b^2-1)}{4k^2+1} \\then~~ 2kb-2k+b^2-1=(2k+b+1)(b-1)=0\to2k+b+1=0 \\overall ~~(2,-1)\quad \blacksquare

  2. normal

    let  A(x1,y1)  B(x2,y2)AB:y=kx+bwe note  that  (x1−2,y1)⋅(x2−2,y2)=0simplify  (k2+1)x1x2+(bk−2)(x1+x2)+4+b2=0x1+x2=−8kb4k2+3x1x2=4(b2−3)4k2+3then  4k2+16kb+7b2=0then  2k+b=0(quit) or 2k+7b=0(27,0)■let~~A(x_1,y_1)~~B(x_2,y_2)\quad AB:y=kx+b \\we~note~~that ~~(x_1-2,y_1)\cdot(x_2-2,y_2)=0 \\simplify~~(k^2+1)x_1x_2+(bk-2)(x_1+x_2)+4+b^2=0 \\x_1+x_2=\frac{-8kb}{4k^2+3}\quad x_1x_2=\frac{4(b^2-3)}{4k^2+3} \\then~~4k^2+16kb+7b^2=0 \\then~~2k+b=0(quit)~or~2k+7b=0 \\(\frac{2}{7},0)\quad \blacksquare

  3. normal

    let  P′(x1,y1)  Q(x2,y2)P′Q:y=kx+bwe  note  that  PQR  collinearthat  means  k(x1−x2)x2+x2=y1−1x1simplify  2kx1x2+(b−1)(x1+x2)=0x1+x2=−8kb4k2+2x1x2=4(b2−2)4k2+2then  k(b−2)=0  then  b=2(0,2)■let~~P'(x_1,y_1)~~Q(x_2,y_2)\quad P'Q:y=kx+b \\we~~note~~that~~PQR~~collinear \\that~~means~~\frac{k(x_1-x_2)}{x_2+x_2}=\frac{y_1-1}{x_1} \\simplify~~2kx_1x_2+(b-1)(x_1+x_2)=0 \\x_1+x_2=\frac{-8kb}{4k^2+2}\quad x_1x_2=\frac{4(b^2-2)}{4k^2+2} \\then~~k(b-2)=0~~then~~b=2 \\(0,2)\quad \blacksquare
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