巧学数学齐次化法联立妙解圆锥曲线问题！
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巧学数学齐次化法联立妙解圆锥曲线问题！
请你多多关注，学习少走弯路，成绩突飞猛进，高考考题全对！齐次化法简化计算适用范围：圆锥曲线中处理斜率之和与斜率之积类型问题写这个专题是因为2017年全国I卷再次考到该类问题，构造齐次处理此类问题已经流行很久，所谓的通性通法不是指自己不熟练的或者是没有研究过的就不是通法，当然下面几个例子都可以由的直线与曲线方程联立消元然后韦达定理的“通法”做出来。
喜欢该文的人也喜欢&p&昨天看到了一道抛物线的斜率问题，发现齐次化处理很容易，而常规做法运算量相当大，与大家分享：&/p&&p&例（17武汉二调）直线y=2x-3与抛物线 &img src=&https://www.zhihu.com/equation?tex=y%5E%7B2%7D%3D4x& alt=&y^{2}=4x& eeimg=&1&& 交于A,B两点，且OA,OB的斜率分别为 &img src=&https://www.zhihu.com/equation?tex=k_%7B1%7D%EF%BC%8Ck_%7B2%7D& alt=&k_{1}，k_{2}& eeimg=&1&& ，求 &img src=&https://www.zhihu.com/equation?tex=%5Cfrac%7B1%7D%7Bk_%7B1%7D%7D%2B%5Cfrac%7B1%7D%7Bk_%7B2%7D%7D& alt=&\frac{1}{k_{1}}+\frac{1}{k_{2}}& eeimg=&1&& 的值&/p&&p&解：即构造含x/y的齐次式，由2x-y=3得 &img src=&https://www.zhihu.com/equation?tex=%5Cfrac%7B2x-y%7D%7B3%7D%3D1& alt=&\frac{2x-y}{3}=1& eeimg=&1&& ，故 &img src=&https://www.zhihu.com/equation?tex=y%5E%7B2%7D%3D4%5Ccdot%5Cfrac%7B2x-y%7D%7B3%7D%5Ccdot+x& alt=&y^{2}=4\cdot\frac{2x-y}{3}\cdot x& eeimg=&1&& ，这样我们就完成了对抛物线的齐次化处理，即：&/p&&p&&img src=&https://www.zhihu.com/equation?tex=y%5E%7B2%7D%3D%5Cfrac%7B8%7D%7B3%7Dx%5E%7B2%7D-%5Cfrac%7B4%7D%7B3%7Dxy%5C%5C& alt=&y^{2}=\frac{8}{3}x^{2}-\frac{4}{3}xy\\& eeimg=&1&&&/p&&p&两边同时除以y^2且令t=x/y得：&/p&&p&&img src=&https://www.zhihu.com/equation?tex=%5Cfrac%7B8%7D%7B3%7Dt%5E%7B2%7D-%5Cfrac%7B4%7D%7B3%7Dt-1%3D0%5C%5C& alt=&\frac{8}{3}t^{2}-\frac{4}{3}t-1=0\\& eeimg=&1&&&/p&&p& 故 &img src=&https://www.zhihu.com/equation?tex=t_%7B1%7D%2Bt_%7B2%7D%3D%5Cfrac%7B1%7D%7B2%7D& alt=&t_{1}+t_{2}=\frac{1}{2}& eeimg=&1&& 即所求值为1/2&/p&&p&&br&&/p&&p&我们将题目一般化可得下面这个漂亮的结论：&/p&&p&&b&对于抛物线 &img src=&https://www.zhihu.com/equation?tex=y%5E%7B2%7D%3D2px& alt=&y^{2}=2px& eeimg=&1&& ，一直线y=kx+b与其交于两点，两点与原点连线斜率的倒数和为定值 &img src=&https://www.zhihu.com/equation?tex=%5Cfrac%7B1%7D%7Bk%7D& alt=&\frac{1}{k}& eeimg=&1&&&/b& &/p&&p&此结论同样使用上面的齐次化方法非常容易证明，请读者自行完成以加深方法与结论的印象&/p&&p&&br&&/p&&p&思考：请读者利用此结论&b&自行命制几道试题&/b&并且思考这个命题的&b&逆命题是否成立&/b&&/p&&p&关于齐次化的基本方法还不清楚的同学可以先看一下我的这篇专栏：&/p&&a href=&https://zhuanlan.zhihu.com/p/& data-draft-node=&block& data-draft-type=&link-card& data-image=&https://pic1.zhimg.com/equation.jpg& data-image-width=&0& data-image-height=&0& class=&internal&&Owl.G：圆锥曲线中的齐次化方法（1）-基本思路&/a&&p&2月28日更新：&/p&&p&刚刚拿到了今天刚考完的武汉市二月调考的试卷，做了一下发现小题第六题和上面这道去年的题目源于同一个背景：&/p&&p&例（18武汉二调）已知不过原点的直线O交抛物线 &img src=&https://www.zhihu.com/equation?tex=y%5E%7B2%7D%3D2px& alt=&y^{2}=2px& eeimg=&1&& 于A,B两点，若 &img src=&https://www.zhihu.com/equation?tex=k_%7BOA%7D%3D2%EF%BC%8Ck_%7BAB%7D%3D6& alt=&k_{OA}=2，k_{AB}=6& eeimg=&1&& ，则 &img src=&https://www.zhihu.com/equation?tex=k_%7BOB%7D& alt=&k_{OB}& eeimg=&1&& 为？&/p&&p&
解：利用上面的一般化结论有 &img src=&https://www.zhihu.com/equation?tex=%5Cfrac%7B1%7D%7Bk_%7BOB%7D%7D%2B%5Cfrac%7B1%7D%7B2%7D%3D%5Cfrac%7B1%7D%7B6%7D& alt=&\frac{1}{k_{OB}}+\frac{1}{2}=\frac{1}{6}& eeimg=&1&& ，得所求为-3&/p&&p&
很快吧嘿嘿～&/p&&p&&/p&&p&&/p&&p&&/p&&p&&/p&
昨天看到了一道抛物线的斜率问题，发现齐次化处理很容易，而常规做法运算量相当大，与大家分享：例（17武汉二调）直线y=2x-3与抛物线 y^{2}=4x 交于A,B两点，且OA,OB的斜率分别为 k_{1}，k_{2} ，求 \frac{1}{k_{1}}+\frac{1}{k_{2}} 的值解：即构造含x/y的…
&figure&&img src=&https://pic4.zhimg.com/v2-79cf35f846f3a52cf4c59b_b.jpg& data-rawwidth=&1920& data-rawheight=&1080& class=&origin_image zh-lightbox-thumb& width=&1920& data-original=&https://pic4.zhimg.com/v2-79cf35f846f3a52cf4c59b_r.jpg&&&/figure&&p&这个是我们的数学老师在课上提到的方法，原本是讲2018年泉州市高三质检的题目，后来发现可以做2017年的全国一卷理数的压轴题，并且特别快！！&/p&&p&&br&&/p&&p&过去尝试用一般的方法做这道题，也就是：设点、设线、求斜率。&/p&&p&计算量非常大，最后能够得到一串表达式。&/p&&p&得到那串式子（或是用我们数学老师的话说，那一坨）后，要代回去求定点 。&/p&&p&用个人最快的运算速度来做的话，要花近十分钟。&/p&&p&&br&&/p&&p&但是用双联立的方法，式子一列，熟练的话一分钟可以解决。&/p&&p&没错！你没有看错！一分钟！！&/p&&p&（在这里向我们的数学老师比心）&/p&&p&&br&&/p&&figure&&img src=&https://pic2.zhimg.com/v2-dcdd324ee1d9d_b.jpg& data-size=&normal& data-rawwidth=&730& data-rawheight=&184& class=&origin_image zh-lightbox-thumb& width=&730& data-original=&https://pic2.zhimg.com/v2-dcdd324ee1d9d_r.jpg&&&figcaption&例题 (2017年全国1理)&/figcaption&&/figure&&p&第一步只要代入 &img src=&https://www.zhihu.com/equation?tex=P_2& alt=&P_2& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=P_3& alt=&P_3& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=P_4& alt=&P_4& eeimg=&1&& 四个点，就可以求出椭圆的方程为 &img src=&https://www.zhihu.com/equation?tex=C%EF%BC%9A%5Cfrac%7Bx%5E2%7D%7B4%7D%2By%5E2%3D1& alt=&C：\frac{x^2}{4}+y^2=1& eeimg=&1&& 。&/p&&p&接下来直接做第二步。&/p&&p&&b&解 &/b&设直线 &img src=&https://www.zhihu.com/equation?tex=P_2A& alt=&P_2A& eeimg=&1&& 的斜率为 &img src=&https://www.zhihu.com/equation?tex=k_1& alt=&k_1& eeimg=&1&& ，直线 &img src=&https://www.zhihu.com/equation?tex=P_2B& alt=&P_2B& eeimg=&1&& 的斜率为 &img src=&https://www.zhihu.com/equation?tex=k_2& alt=&k_2& eeimg=&1&& ，&/p&&p&&img src=&https://www.zhihu.com/equation?tex=P_2A%3Ay%3Dk_1x%2B1& alt=&P_2A:y=k_1x+1& eeimg=&1&&，即 &img src=&https://www.zhihu.com/equation?tex=k_1x%2B%281-y%29%3D0& alt=&k_1x+(1-y)=0& eeimg=&1&&&/p&&p&同理 &img src=&https://www.zhihu.com/equation?tex=P_2B%EF%BC%9Ak_2x%2B%281-y%29%3D0& alt=&P_2B：k_2x+(1-y)=0& eeimg=&1&&&/p&&p&此时将 &img src=&https://www.zhihu.com/equation?tex=P_2A& alt=&P_2A& eeimg=&1&& 和 &img src=&https://www.zhihu.com/equation?tex=P_2B& alt=&P_2B& eeimg=&1&& 的方程合在一起： &img src=&https://www.zhihu.com/equation?tex=%5Bk_1x%2B%281-y%29%5D%5Ccdot+%5Bk_2x%2B%281-y%29%5D%3D0& alt=&[k_1x+(1-y)]\cdot [k_2x+(1-y)]=0& eeimg=&1&&&/p&&p&这个式子就可以表示 &img src=&https://www.zhihu.com/equation?tex=P_2A& alt=&P_2A& eeimg=&1&& 和 &img src=&https://www.zhihu.com/equation?tex=P_2B& alt=&P_2B& eeimg=&1&& 这两个直线上所有的点，化简后得到&/p&&p&&img src=&https://www.zhihu.com/equation?tex=k_1k_2x%5E2%2B%28k_1%2Bk_2%29x%281-y%29%2B%281-y%29%5E2%3D0& alt=&k_1k_2x^2+(k_1+k_2)x(1-y)+(1-y)^2=0& eeimg=&1&&&/p&&p&题目条件有 &img src=&https://www.zhihu.com/equation?tex=k_1%2Bk_2%3D-1& alt=&k_1+k_2=-1& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=%5Cfrac%7Bx%5E2%7D%7B4%7D%2By%5E2%3D1& alt=&\frac{x^2}{4}+y^2=1& eeimg=&1&&&/p&&p&即 &img src=&https://www.zhihu.com/equation?tex=x%5E2%3D4%281-y%5E2%29%3D4%281%2By%29%281-y%29& alt=&x^2=4(1-y^2)=4(1+y)(1-y)& eeimg=&1&&&/p&&p&代入得 &img src=&https://www.zhihu.com/equation?tex=4k_1k_2%281%2By%29%281-y%29-x%281-y%29%2B%281-y%29%5E2%3D0& alt=&4k_1k_2(1+y)(1-y)-x(1-y)+(1-y)^2=0& eeimg=&1&&&/p&&p&因为直线 &img src=&https://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&& 不经过 &img src=&https://www.zhihu.com/equation?tex=P_2%280%2C1%29& alt=&P_2(0,1)& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=y%5Cne1& alt=&y\ne1& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=1-y%5Cne0& alt=&1-y\ne0& eeimg=&1&&&/p&&p&所以消去 &img src=&https://www.zhihu.com/equation?tex=%281-y%29& alt=&(1-y)& eeimg=&1&& 得&img src=&https://www.zhihu.com/equation?tex=l%3A4k_1k_2%281%2By%29-x%2B%281-y%29%3D0& alt=&l:4k_1k_2(1+y)-x+(1-y)=0& eeimg=&1&& 。&/p&&p&在这里，要求出直线所过的定点，只需联立以下方程：&/p&&p&&img src=&https://www.zhihu.com/equation?tex=1%2By%3D0& alt=&1+y=0& eeimg=&1&& 和 &img src=&https://www.zhihu.com/equation?tex=-x%2B%281-y%29%3D0& alt=&-x+(1-y)=0& eeimg=&1&& ，解得 &img src=&https://www.zhihu.com/equation?tex=x%3D2& alt=&x=2& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=y%3D-1& alt=&y=-1& eeimg=&1&&&/p&&p&即直线 &img src=&https://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&& 过定点 &img src=&https://www.zhihu.com/equation?tex=%282%2C-1%29& alt=&(2,-1)& eeimg=&1&&&/p&&p&&br&&/p&&p&换个角度，我们再来做一下这道题，现在跳过无关的讲解，只写关键步骤。&/p&&p&&br&&/p&&p&&b&例&/b& 若直线 &img src=&https://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&& 经过定点 &img src=&https://www.zhihu.com/equation?tex=%282%2C-1%29& alt=&(2,-1)& eeimg=&1&& ，且不经过点 &img src=&https://www.zhihu.com/equation?tex=P_2%280%2C1%29& alt=&P_2(0,1)& eeimg=&1&& ,与 &img src=&https://www.zhihu.com/equation?tex=C& alt=&C& eeimg=&1&& 交于 &img src=&https://www.zhihu.com/equation?tex=A& alt=&A& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=B& alt=&B& eeimg=&1&& 两点，证明：直线 &img src=&https://www.zhihu.com/equation?tex=P_2A& alt=&P_2A& eeimg=&1&& 与直线 &img src=&https://www.zhihu.com/equation?tex=P_2B& alt=&P_2B& eeimg=&1&& 的斜率的和为定值．&/p&&p&&b&解&/b& 设&img src=&https://www.zhihu.com/equation?tex=P_2A%3Ay%3Dk_1x%2B1& alt=&P_2A:y=k_1x+1& eeimg=&1&&，即 &img src=&https://www.zhihu.com/equation?tex=k_1x%2B%281-y%29%3D0& alt=&k_1x+(1-y)=0& eeimg=&1&&，&/p&&p&同理 &img src=&https://www.zhihu.com/equation?tex=P_2B%EF%BC%9Ak_2x%2B%281-y%29%3D0& alt=&P_2B：k_2x+(1-y)=0& eeimg=&1&&&/p&&p&&img src=&https://www.zhihu.com/equation?tex=%5Bk_1x%2B%281-y%29%5D%5Ccdot+%5Bk_2x%2B%281-y%29%5D%3D0& alt=&[k_1x+(1-y)]\cdot [k_2x+(1-y)]=0& eeimg=&1&&&/p&&p&化简得： &img src=&https://www.zhihu.com/equation?tex=k_1k_2x%5E2%2B%28k_1%2Bk_2%29x%281-y%29%2B%281-y%29%5E2%3D0& alt=&k_1k_2x^2+(k_1+k_2)x(1-y)+(1-y)^2=0& eeimg=&1&&&/p&&p&因为 &img src=&https://www.zhihu.com/equation?tex=%5Cfrac%7Bx%5E2%7D%7B4%7D%2By%5E2%3D1& alt=&\frac{x^2}{4}+y^2=1& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=x%5E2%3D4%281-y%5E2%29%3D4%281%2By%29%281-y%29& alt=&x^2=4(1-y^2)=4(1+y)(1-y)& eeimg=&1&&&/p&&p&所以 &img src=&https://www.zhihu.com/equation?tex=4k_1k_2%281%2By%29%281-y%29%2B%28k_1%2Bk_2%29x%281-y%29%2B%281-y%29%5E2%3D0& alt=&4k_1k_2(1+y)(1-y)+(k_1+k_2)x(1-y)+(1-y)^2=0& eeimg=&1&&&/p&&p&因为 &img src=&https://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&& 不经过点 &img src=&https://www.zhihu.com/equation?tex=P_2& alt=&P_2& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=y%5Cne1& alt=&y\ne1& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=1-y%5Cne0& alt=&1-y\ne0& eeimg=&1&& ，&/p&&p&所以消去 &img src=&https://www.zhihu.com/equation?tex=%281-y%29& alt=&(1-y)& eeimg=&1&& 得&img src=&https://www.zhihu.com/equation?tex=l%3A4k_1k_2%281%2By%29%2B%28k_1%2Bk_2%29x%2B%281-y%29%3D0& alt=&l:4k_1k_2(1+y)+(k_1+k_2)x+(1-y)=0& eeimg=&1&&&/p&&p&因为 &img src=&https://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&& 经过定点 &img src=&https://www.zhihu.com/equation?tex=%282%2C-1%29& alt=&(2,-1)& eeimg=&1&& ，代入得&/p&&p&&img src=&https://www.zhihu.com/equation?tex=2%28k_1%2Bk_2%29%2B2%3D0& alt=&2(k_1+k_2)+2=0& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=k_1%2Bk_2%3D-1& alt=&k_1+k_2=-1& eeimg=&1&&&/p&&p&&br&&/p&&p&注意看上面的两个例子可以发现：计算过程中都正好可以消去一个值。要是没办法消去的话，是不是就不能用这个方法了？&/p&&p&下面给出一个例子，题目来源自 &a class=&member_mention& href=&https://www.zhihu.com/people/0e21fb11fe85e6a6c9a83& data-hash=&0e21fb11fe85e6a6c9a83& data-hovercard=&p$b$0e21fb11fe85e6a6c9a83&&@秦子淳&/a&：&/p&&p&&b&例 &/b&已知椭圆 &img src=&https://www.zhihu.com/equation?tex=%5Cfrac%7Bx%5E2%7D%7B4%7D%2B%5Cfrac%7By%5E2%7D%7B3%7D%3D1& alt=&\frac{x^2}{4}+\frac{y^2}{3}=1& eeimg=&1&& 的右顶点 &img src=&https://www.zhihu.com/equation?tex=P%282%2C0%29& alt=&P(2,0)& eeimg=&1&& ，过右顶点作两条相互垂直的直线交椭圆于 &img src=&https://www.zhihu.com/equation?tex=A%2CB& alt=&A,B& eeimg=&1&& 两点，求证：直线 &img src=&https://www.zhihu.com/equation?tex=AB& alt=&AB& eeimg=&1&& 过定点。&/p&&p&&b&解&/b& 设 &img src=&https://www.zhihu.com/equation?tex=PA%3Ay%3Dk_1%28x-2%29& alt=&PA:y=k_1(x-2)& eeimg=&1&& ，即 &img src=&https://www.zhihu.com/equation?tex=k_1%28x-2%29-y%3D0& alt=&k_1(x-2)-y=0& eeimg=&1&&&/p&&p&同理 &img src=&https://www.zhihu.com/equation?tex=PB%3Ak_2%28x-2%29-y%3D0& alt=&PB:k_2(x-2)-y=0& eeimg=&1&&&/p&&p&&img src=&https://www.zhihu.com/equation?tex=%5Bk_1%28x-2%29-y%5D%5Ccdot%5Bk_2%28x-2%29-y%5D%3D0& alt=&[k_1(x-2)-y]\cdot[k_2(x-2)-y]=0& eeimg=&1&&&/p&&p&&img src=&https://www.zhihu.com/equation?tex=k_1k_2%28x-2%29%5E2-%28k_1%2Bk_2%29%28x-2%29y%2By%5E2%3D0& alt=&k_1k_2(x-2)^2-(k_1+k_2)(x-2)y+y^2=0& eeimg=&1&&&/p&&p&由于两直线互相垂直， &img src=&https://www.zhihu.com/equation?tex=k_1k_2%3D-1& alt=&k_1k_2=-1& eeimg=&1&& ，并且注意到旁边“孤单”的 &img src=&https://www.zhihu.com/equation?tex=y%5E2& alt=&y^2& eeimg=&1&& ，&/p&&p&用&img src=&https://www.zhihu.com/equation?tex=y%5E2%3D3-%5Cfrac%7B3%7D%7B4%7Dx%5E2& alt=&y^2=3-\frac{3}{4}x^2& eeimg=&1&& 消去 &img src=&https://www.zhihu.com/equation?tex=y%5E2& alt=&y^2& eeimg=&1&& 得：&/p&&p&&img src=&https://www.zhihu.com/equation?tex=-%28x%5E2-4x%2B4%29-%28k_1%2Bk_2%29%28x-2%29y%2B3-%5Cfrac%7B3%7D%7B4%7Dx%5E2%3D0& alt=&-(x^2-4x+4)-(k_1+k_2)(x-2)y+3-\frac{3}{4}x^2=0& eeimg=&1&&&/p&&p&整理得：&/p&&p&&img src=&https://www.zhihu.com/equation?tex=%28-%5Cfrac%7B7%7D%7B4%7Dx%5E2%2B4x-1%29-%28k_1%2Bk_2%29%28x-2%29y%3D0& alt=&(-\frac{7}{4}x^2+4x-1)-(k_1+k_2)(x-2)y=0& eeimg=&1&&&/p&&p&联立方程 &img src=&https://www.zhihu.com/equation?tex=-%5Cfrac%7B7%7D%7B4%7Dx%5E2%2B4x-1%3D0& alt=&-\frac{7}{4}x^2+4x-1=0& eeimg=&1&& 和 &img src=&https://www.zhihu.com/equation?tex=%28x-2%29y%3D0& alt=&(x-2)y=0& eeimg=&1&& 得&/p&&p&&img src=&https://www.zhihu.com/equation?tex=x_1%3D2& alt=&x_1=2& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=x_2%3D%5Cfrac%7B2%7D%7B7%7D& alt=&x_2=\frac{2}{7}& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=y%3D0& alt=&y=0& eeimg=&1&&&/p&&p&所以过定点 &img src=&https://www.zhihu.com/equation?tex=%28%5Cfrac%7B2%7D%7B7%7D%2C0%29& alt=&(\frac{2}{7},0)& eeimg=&1&&&/p&&p&&br&&/p&&p&在做完这几道题后，我们可以得到如下框架：&/p&&figure&&img src=&https://pic2.zhimg.com/v2-f3aee2ab5b7ef5de0a6dd97_b.jpg& data-size=&normal& data-rawwidth=&248& data-rawheight=&210& class=&content_image& width=&248&&&figcaption&已知两直线斜率的和或积可以推出直线过定点；已知直线过定点可以推出两直线的和或积&/figcaption&&/figure&&p&当然，虽然这个方法很棒，但是书写过程要特别注意，并且不是所有时候都可以算出答案。&/p&&p&放上我们数学老师的原话：这个方法慎用。&/p&&p&评论区有人问到不能用的情形，在这里举个例子：&/p&&p&椭圆 &img src=&https://www.zhihu.com/equation?tex=%5Cfrac%7Bx%5E2%7D%7B4%7D%2B%5Cfrac%7By%5E2%7D%7B3%7D%3D1& alt=&\frac{x^2}{4}+\frac{y^2}{3}=1& eeimg=&1&& ，过 &img src=&https://www.zhihu.com/equation?tex=A%281%2C1%29& alt=&A(1,1)& eeimg=&1&& 作两条直线 &img src=&https://www.zhihu.com/equation?tex=MN& alt=&MN& eeimg=&1&& 、 &img src=&https://www.zhihu.com/equation?tex=PQ& alt=&PQ& eeimg=&1&& 交椭圆于 &img src=&https://www.zhihu.com/equation?tex=M%EF%BC%8CN%EF%BC%8CP%EF%BC%8CQ& alt=&M，N，P，Q& eeimg=&1&& 四点，两直线斜率满足： &img src=&https://www.zhihu.com/equation?tex=k_1%2Bk_2%3D1& alt=&k_1+k_2=1& eeimg=&1&& ， &img src=&https://www.zhihu.com/equation?tex=B& alt=&B& eeimg=&1&& 和 &img src=&https://www.zhihu.com/equation?tex=C& alt=&C& eeimg=&1&& 分别是 &img src=&https://www.zhihu.com/equation?tex=MN& alt=&MN& eeimg=&1&& 和 &img src=&https://www.zhihu.com/equation?tex=PQ& alt=&PQ& eeimg=&1&& 的中点，求证：直线 &img src=&https://www.zhihu.com/equation?tex=BC& alt=&BC& eeimg=&1&&过定点。&/p&&p&这道题难度极大，在这里不附上解答了，欢迎尝试解答~&/p&&p&&br&&/p&&p&注：和超人老师的某节课有点相似的地方，“齐次化处理”这个名字也是从超人老师那里借鉴过来的，可以当作是超人老师的课的一点补充。&/p&
这个是我们的数学老师在课上提到的方法，原本是讲2018年泉州市高三质检的题目，后来发现可以做2017年的全国一卷理数的压轴题，并且特别快！！ 过去尝试用一般的方法做这道题，也就是：设点、设线、求斜率。计算量非常大，最后能够得到一串表达式。得到那串…
&figure&&img src=&https://pic2.zhimg.com/v2-bcf_b.jpg& data-rawwidth=&1125& data-rawheight=&1080& class=&origin_image zh-lightbox-thumb& width=&1125& data-original=&https://pic2.zhimg.com/v2-bcf_r.jpg&&&/figure&&h2&&ul&&li&&b&引言&/b&&/li&&/ul&&/h2&&p&我们知道，在常见的解析几何问题中，通常问题的解法都包含&b&联立&img src=&http://www.zhihu.com/equation?tex=%5Crightarrow+& alt=&\rightarrow & eeimg=&1&&伟达定理&/b&的步骤，即给定椭圆方程&img src=&http://www.zhihu.com/equation?tex=C& alt=&C& eeimg=&1&&：&img src=&http://www.zhihu.com/equation?tex=%5Cbegin%7Bequation%7D%0A%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cdfrac%7By%5E2%7D%7Bb%5E2%7D%3D1%0A%5Cend%7Bequation%7D& alt=&\begin{equation}
\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1
\end{equation}& eeimg=&1&&&br&&/p&&p&和直线方程&img src=&http://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&&：&img src=&http://www.zhihu.com/equation?tex=y%3Dkx%2Bm& alt=&y=kx+m& eeimg=&1&&&br&&/p&&p&通过&b&联立&/b&获得&/p&&img src=&http://www.zhihu.com/equation?tex=%5Cdfrac%7Bb%5E2%2Ba%5E2k%5E2%7D%7Ba%5E2b%5E2%7Dx%5E2%2B%5Cdfrac%7B2km%7D%7Bb%5E2%7Dx%2B%5Cdfrac%7Bm%5E2-b%5E2%7D%7Bb%5E2%7D%3D0& alt=&\dfrac{b^2+a^2k^2}{a^2b^2}x^2+\dfrac{2km}{b^2}x+\dfrac{m^2-b^2}{b^2}=0& eeimg=&1&&&br&&p&再通过&b&韦达定理&/b&就可以获得两根之间的关系&/p&&img src=&http://www.zhihu.com/equation?tex=%5Cleft%5C%7B+%0A%5Cbegin%7Barray%7D%7Bl%7D%0Ax_1%2Bx_2%3D-%5Cdfrac%7B2a%5E2km%7D%7Bb%5E2%2Ba%5E2k%5E2%7D%5C%5C%0Ax_1x_2%3D%5Cdfrac%7Ba%5E2%28m%5E2-b%5E2%29%7D%7Bb%5E2%2Ba%5E2k%5E2%7D%0A%5Cend%7Barray%7D+%5Cright%5C%7D+& alt=&\left\{
\begin{array}{l}
x_1+x_2=-\dfrac{2a^2km}{b^2+a^2k^2}\\
x_1x_2=\dfrac{a^2(m^2-b^2)}{b^2+a^2k^2}
\end{array} \right\} & eeimg=&1&&&br&&p&并能依据上式求解&img src=&http://www.zhihu.com/equation?tex=y_1%2By_2& alt=&y_1+y_2& eeimg=&1&&和&img src=&http://www.zhihu.com/equation?tex=y_1y_2& alt=&y_1y_2& eeimg=&1&&，利用这些等式去求解问题。&/p&&br&&p&但是对待斜率问题时，我们如果依然按照往常的思路来计算，则计算较为繁琐，这里如果我们能够将斜率&img src=&http://www.zhihu.com/equation?tex=k& alt=&k& eeimg=&1&&看做二次方程中的自变量，那么就不需要像通常我们所做的一般，利用&img src=&http://www.zhihu.com/equation?tex=x_1%2Cx_2& alt=&x_1,x_2& eeimg=&1&&表示斜率再计算，而可以直接求解斜率，从而解决斜率相关的问题。&br&&/p&&br&&h2&&ul&&li&&b&方法介绍&/b&&/li&&/ul&&/h2&&br&&p&我们这里依然有引言中的方程&img src=&http://www.zhihu.com/equation?tex=C%2Cl& alt=&C,l& eeimg=&1&&，现在我们考虑联立，但联立之前我们先对直线方程做一点小处理：&/p&&img src=&http://www.zhihu.com/equation?tex=y%3Dkx%2Bm%5Crightarrow+%5Cdfrac%7By-kx%7D%7Bm%7D%3D1& alt=&y=kx+m\rightarrow \dfrac{y-kx}{m}=1& eeimg=&1&&&br&&p&将处理后的直线方程带入椭圆方程&img src=&http://www.zhihu.com/equation?tex=C& alt=&C& eeimg=&1&&&/p&&p&&img src=&http://www.zhihu.com/equation?tex=%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cdfrac%7By%5E2%7D%7Bb%5E2%7D%3D%5Cleft%28+%5Cdfrac%7By-kx%7D%7Bm%7D+%5Cright%29+%5E2& alt=&\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\left( \dfrac{y-kx}{m} \right) ^2& eeimg=&1&&(将直线方程带入“1”)&br&&/p&&p&展开，得到&/p&&img src=&http://www.zhihu.com/equation?tex=%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cdfrac%7By%5E2%7D%7Bb%5E2%7D%3D%5Cdfrac%7By%5E2-2kxy%2Bk%5E2x%5E2%7D%7Bm%5E2%7D& alt=&\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{y^2-2kxy+k^2x^2}{m^2}& eeimg=&1&&&br&&p&再两边同除&img src=&http://www.zhihu.com/equation?tex=x%5E2& alt=&x^2& eeimg=&1&&，移项&/p&&img src=&http://www.zhihu.com/equation?tex=%5Cleft%28%5Cdfrac%7B1%7D%7Bb%5E2%7D-%5Cdfrac%7B1%7D%7Bm%5E2%7D%5Cright%29%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%5E2%2B%5Cdfrac%7B2k%7D%7Bm%5E2%7D%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7B1%7D%7Ba%5E2%7D-%5Cdfrac%7Bk%5E2%7D%7Bm%5E2%7D%5Cright%29%3D0& alt=&\left(\dfrac{1}{b^2}-\dfrac{1}{m^2}\right)\left(\dfrac{y}{x}\right)^2+\dfrac{2k}{m^2}\left(\dfrac{y}{x}\right)+\left(\dfrac{1}{a^2}-\dfrac{k^2}{m^2}\right)=0& eeimg=&1&&&br&&p&现在，我么就得到了关于&b&变量&img src=&http://www.zhihu.com/equation?tex=%5Cdfrac%7By%7D%7Bx%7D& alt=&\dfrac{y}{x}& eeimg=&1&&的二次方程&/b&，记该方程的两个解分别为&img src=&http://www.zhihu.com/equation?tex=k_1%2Ck_2& alt=&k_1,k_2& eeimg=&1&&与引言类似，这里也可以通过伟达定理得出&/p&&img src=&http://www.zhihu.com/equation?tex=%5Cleft%5C%7B+%0A%5Cbegin%7Barray%7D%7Bl%7D%0Ak_1%2Bk_2%3D-%5Cdfrac%7B2b%5E2k%7D%7Bm%5E2-b%5E2%7D%5C%5C%0Ak_1k_2%3D%5Cdfrac%7Bb%5E2%28m%5E2-a%5E2k%5E2%29%7D%7Ba%5E2%28m%5E2-b%5E2%29%7D%0A%5Cend%7Barray%7D+%5Cright%5C%7D+& alt=&\left\{
\begin{array}{l}
k_1+k_2=-\dfrac{2b^2k}{m^2-b^2}\\
k_1k_2=\dfrac{b^2(m^2-a^2k^2)}{a^2(m^2-b^2)}
\end{array} \right\} & eeimg=&1&&&br&&br&&h2&&ul&&li&&b&具体应用；例题&/b&&/li&&/ul&&/h2&&b&[例1]&/b&已知椭圆&img src=&http://www.zhihu.com/equation?tex=C& alt=&C& eeimg=&1&&：&img src=&http://www.zhihu.com/equation?tex=%5Cdfrac%7Bx%5E2%7D%7B4%7D%2By%5E2%3D1& alt=&\dfrac{x^2}{4}+y^2=1& eeimg=&1&&&br&&p&与直线&img src=&http://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&&：&img src=&http://www.zhihu.com/equation?tex=y%3Dkx%2Bm%28k%5Cne0%29& alt=&y=kx+m(k\ne0)& eeimg=&1&&&/p&&br&&p&交于&img src=&http://www.zhihu.com/equation?tex=M%28x_1%2Cy_1%29%2CN%28x_2%2Cy_2%29& alt=&M(x_1,y_1),N(x_2,y_2)& eeimg=&1&&两点，记&img src=&http://www.zhihu.com/equation?tex=k_%7BOM%7D%3Dk_1%3Bk_%7BON%7D%3Dk_2& alt=&k_{OM}=k_1;k_{ON}=k_2& eeimg=&1&&&/p&&p&试证明当&img src=&http://www.zhihu.com/equation?tex=3%28k_1%2Bk_2%29%3D8k& alt=&3(k_1+k_2)=8k& eeimg=&1&&时，&img src=&http://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&&过一定点。&/p&&br&&b&齐次化解法&/b&：&p&处理直线方程&img src=&http://www.zhihu.com/equation?tex=y%3Dkx%2Bm%5Crightarrow+%5Cdfrac%7By-kx%7D%7Bm%7D%3D1& alt=&y=kx+m\rightarrow \dfrac{y-kx}{m}=1& eeimg=&1&&&/p&&p&&b&联立&/b&可得&img src=&http://www.zhihu.com/equation?tex=%5Cleft%281-%5Cdfrac%7B1%7D%7Bm%5E2%7D%5Cright%29%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%5E2%2B%5Cdfrac%7B2k%7D%7Bm%5E2%7D%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D-%5Cdfrac%7Bk%5E2%7D%7Bm%5E2%7D%5Cright%29%3D0& alt=&\left(1-\dfrac{1}{m^2}\right)\left(\dfrac{y}{x}\right)^2+\dfrac{2k}{m^2}\left(\dfrac{y}{x}\right)+\left(\dfrac{1}{4}-\dfrac{k^2}{m^2}\right)=0& eeimg=&1&&&br&&/p&&p&进而由&b&韦达定理&/b&可得&img src=&http://www.zhihu.com/equation?tex=%5Cleft%5C%7B+%0A%5Cbegin%7Barray%7D%7Bl%7D%0Ak_1%2Bk_2%3D-%5Cdfrac%7B2k%7D%7Bm%5E2-1%7D%5C%5C%0Ak_1k_2%3D%5Cdfrac%7Bm%5E2-4k%5E2%7D%7B4%28m%5E2-1%29%7D%0A%5Cend%7Barray%7D+%5Cright%5C%7D+& alt=&\left\{
\begin{array}{l}
k_1+k_2=-\dfrac{2k}{m^2-1}\\
k_1k_2=\dfrac{m^2-4k^2}{4(m^2-1)}
\end{array} \right\} & eeimg=&1&&&br&&/p&&p&结合题目条件&/p&&img src=&http://www.zhihu.com/equation?tex=3%28k_1%2Bk_2%29%3D8k& alt=&3(k_1+k_2)=8k& eeimg=&1&&&br&&p&可得&img src=&http://www.zhihu.com/equation?tex=3%28k_1%2Bk_2%29%3D-%5Cdfrac%7B6k%7D%7Bm%5E2-1%7D%3D8k%5CRightarrow+m%3D%5Cpm+%5Cdfrac%7B1%7D%7B2%7D& alt=&3(k_1+k_2)=-\dfrac{6k}{m^2-1}=8k\Rightarrow m=\pm \dfrac{1}{2}& eeimg=&1&&&br&&/p&&p&所以直线&img src=&http://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&&必过定点&img src=&http://www.zhihu.com/equation?tex=%5Cleft%280%2C%5Cpm%5Cdfrac%7B1%7D%7B2%7D%5Cright%29& alt=&\left(0,\pm\dfrac{1}{2}\right)& eeimg=&1&&&/p&&br&&p&现在，我们已经对关于原点斜率问题有了处理方法，接下来我们考虑解决不关于原点的斜率的问题，处理方法是类似的，将其转化为过原点的斜率问题即可，具体处理方式请看下例：&/p&&br&&p&&b&[例2](不过原点的情况)&/b&已知椭圆方程&img src=&http://www.zhihu.com/equation?tex=C& alt=&C& eeimg=&1&&：&img src=&http://www.zhihu.com/equation?tex=%5Cdfrac%7Bx%5E2%7D%7B4%7D%2B%5Cdfrac%7By%5E2%7D%7B3%7D%3D1& alt=&\dfrac{x^2}{4}+\dfrac{y^2}{3}=1& eeimg=&1&&&br&&/p&&p&与直线&img src=&http://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&&交于&img src=&http://www.zhihu.com/equation?tex=A%28x_1%2Cy_1%29%2CB%28x_2%2Cy_2%29& alt=&A(x_1,y_1),B(x_2,y_2)& eeimg=&1&&两点(&img src=&http://www.zhihu.com/equation?tex=A%2CB& alt=&A,B& eeimg=&1&&不为左右顶点)&br&&/p&&p&以&img src=&http://www.zhihu.com/equation?tex=A%2CB& alt=&A,B& eeimg=&1&&为直径的圆恒过椭圆的左顶点。&br&&/p&&p&求证：直线过定点，并求出该定点。&/p&&p&&b&齐次化解法：&/b&&br&&/p&&p&考虑平移椭圆，目的是使左顶点成为新坐标系的“原点”，根据“左加右减”的原则，将该图像向右平移两格有&/p&&p&&img src=&http://www.zhihu.com/equation?tex=C%27& alt=&C'& eeimg=&1&&：&img src=&http://www.zhihu.com/equation?tex=%5Cdfrac%7B%28x-2%29%5E2%7D%7B4%7D%2B%5Cdfrac%7By%5E2%7D%7B3%7D%3D1%5CLeftrightarrow+%5Cdfrac%7Bx%5E2%7D%7B4%7D%2B%5Cdfrac%7By%5E2%7D%7B3%7D-x%3D0& alt=&\dfrac{(x-2)^2}{4}+\dfrac{y^2}{3}=1\Leftrightarrow \dfrac{x^2}{4}+\dfrac{y^2}{3}-x=0& eeimg=&1&&&br&&/p&&p&现在设直线&img src=&http://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&&:&img src=&http://www.zhihu.com/equation?tex=y%3Dkx%2Bm%5CLeftrightarrow+%5Cdfrac%7By-kx%7D%7Bm%7D%3D1& alt=&y=kx+m\Leftrightarrow \dfrac{y-kx}{m}=1& eeimg=&1&&&/p&&p&现在通过平移图像，我们已经初步转化为了之前说明过的情况，但是椭圆方程中的&1&不见了，而且出现了一次项。所以接下来联立方式略有不同。&b&联立：&/b&&/p&&p&&img src=&http://www.zhihu.com/equation?tex=%5Cdfrac%7Bx%5E2%7D%7B4%7D%2B%5Cdfrac%7By%5E2%7D%7B3%7D-%5Cdfrac%7By-kx%7D%7Bm%7Dx%3D0%5CRightarrow+%5Cdfrac%7B1%7D%7B3%7D%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%5E2-%5Cdfrac%7B1%7D%7Bm%7D%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%2B%5Cdfrac%7Bm%2B4k%7D%7B4m%7D%3D0& alt=&\dfrac{x^2}{4}+\dfrac{y^2}{3}-\dfrac{y-kx}{m}x=0\Rightarrow \dfrac{1}{3}\left(\dfrac{y}{x}\right)^2-\dfrac{1}{m}\left(\dfrac{y}{x}\right)+\dfrac{m+4k}{4m}=0& eeimg=&1&&&br&(这里将&img src=&http://www.zhihu.com/equation?tex=x& alt=&x& eeimg=&1&&看做&img src=&http://www.zhihu.com/equation?tex=1%5Ccdot+x& alt=&1\cdot x& eeimg=&1&&再作&1&的代换)&br&&/p&&p&而&/p&&blockquote&以&img src=&http://www.zhihu.com/equation?tex=A%2CB& alt=&A,B& eeimg=&1&&为直径的圆横过椭圆的左顶点。&/blockquote&&p&事实上现在等价于&img src=&http://www.zhihu.com/equation?tex=k_%7BOA%7D%5Ccdot+k_%7BOB%7D%3D-1& alt=&k_{OA}\cdot k_{OB}=-1& eeimg=&1&&&/p&&p&接下来处理方式与&b&[例1]&/b&完全类似，可以得到&br&&/p&&img src=&http://www.zhihu.com/equation?tex=k_1k_2%3D%5Cdfrac%7B3%28m%2B4k%29%7D%7B4m%7D%3D-1%5CRightarrow+7m%2B12k%3D0%5CRightarrow+m%3D-%5Cfrac%7B12%7D%7B7%7Dk& alt=&k_1k_2=\dfrac{3(m+4k)}{4m}=-1\Rightarrow 7m+12k=0\Rightarrow m=-\frac{12}{7}k& eeimg=&1&&&br&&p&所以直线：&img src=&http://www.zhihu.com/equation?tex=y%3Dk%5Cleft%28x-%5Cdfrac%7B12%7D%7B7%7D%5Cright%29& alt=&y=k\left(x-\dfrac{12}{7}\right)& eeimg=&1&&过定点&img src=&http://www.zhihu.com/equation?tex=%5Cleft%28%5Cdfrac%7B12%7D%7B7%7D%2C+0%5Cright%29& alt=&\left(\dfrac{12}{7}, 0\right)& eeimg=&1&&&/p&&p&当然，要记住我们之前平移了坐标系，现在的定点也是相对于新的原点(事实上原来的左顶点)而言的，所以在初始坐标系中，该定点的坐标为&img src=&http://www.zhihu.com/equation?tex=%5Cleft%28-%5Cdfrac%7B2%7D%7B7%7D%2C0%5Cright%29& alt=&\left(-\dfrac{2}{7},0\right)& eeimg=&1&&&/p&&p&关于联立方式不同，事实上是为了保证方程的&b&齐次性(即可以通过同除&img src=&http://www.zhihu.com/equation?tex=x%5E2& alt=&x^2& eeimg=&1&&得到关于斜率&img src=&http://www.zhihu.com/equation?tex=k& alt=&k& eeimg=&1&&的方程)&/b&，为保证齐次性，我们要求各项的次数都是2，所以我们把一次项的&1&换成了&img src=&http://www.zhihu.com/equation?tex=%5Cdfrac%7By-kx%7D%7Bm%7D& alt=&\dfrac{y-kx}{m}& eeimg=&1&&,二次项的系数就不需要变动，而对于常数项，想要保证其也成为2次项，就需要将其中的&1&换成&img src=&http://www.zhihu.com/equation?tex=%5Cleft%28%5Cdfrac%7By-kx%7D%7Bm%7D%5Cright%29%5E2& alt=&\left(\dfrac{y-kx}{m}\right)^2& eeimg=&1&&。&br&&/p&&br&&h2&&ul&&li&&b&一道练习题&/b&&/li&&/ul&&/h2&椭圆&img src=&http://www.zhihu.com/equation?tex=C& alt=&C& eeimg=&1&&：&img src=&http://www.zhihu.com/equation?tex=%5Cbegin%7Bequation%7D%0A%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cdfrac%7By%5E2%7D%7Bb%5E2%7D%3D1%0A%5Cend%7Bequation%7D& alt=&\begin{equation}
\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1
\end{equation}& eeimg=&1&&&br&&p&与直线&img src=&http://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&&：&img src=&http://www.zhihu.com/equation?tex=y%3Dkx%2Bm%28k%5Cne0%29& alt=&y=kx+m(k\ne0)& eeimg=&1&&&/p&&p&交于&img src=&http://www.zhihu.com/equation?tex=A%28x_1%2Cy_1%29%2CB%28x_2%2Cy_2%29& alt=&A(x_1,y_1),B(x_2,y_2)& eeimg=&1&&两点，以&img src=&http://www.zhihu.com/equation?tex=A%2CB& alt=&A,B& eeimg=&1&&为直径的圆经过原点。&/p&&p&证明：原点到直线&img src=&http://www.zhihu.com/equation?tex=l& alt=&l& eeimg=&1&&的距离是定值。&/p&&p&&b&解答&/b&(大概也能当个例题？)&b&：&/b&&/p&&p&对直线方程作处理后直接&b&联立：&/b&&/p&&img src=&http://www.zhihu.com/equation?tex=%5Cleft%28%5Cdfrac%7B1%7D%7Bb%5E2%7D-%5Cdfrac%7B1%7D%7Bm%5E2%7D%5Cright%29%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%5E2%2B%5Cdfrac%7B2k%7D%7Bm%5E2%7D%5Cleft%28%5Cdfrac%7By%7D%7Bx%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7B1%7D%7Ba%5E2%7D-%5Cdfrac%7Bk%5E2%7D%7Bm%5E2%7D%5Cright%29%3D0& alt=&\left(\dfrac{1}{b^2}-\dfrac{1}{m^2}\right)\left(\dfrac{y}{x}\right)^2+\dfrac{2k}{m^2}\left(\dfrac{y}{x}\right)+\left(\dfrac{1}{a^2}-\dfrac{k^2}{m^2}\right)=0& eeimg=&1&&&br&&p&&b&韦达定理：&/b&&/p&&img src=&http://www.zhihu.com/equation?tex=%5Cleft%5C%7B+%0A%5Cbegin%7Barray%7D%7Bl%7D%0Ak_1%2Bk_2%3D-%5Cdfrac%7B2b%5E2k%7D%7Bm%5E2-b%5E2%7D%5C%5C%0Ak_1k_2%3D%5Cdfrac%7Bb%5E2%28m%5E2-a%5E2k%5E2%29%7D%7Ba%5E2%28m%5E2-b%5E2%29%7D%0A%5Cend%7Barray%7D+%5Cright%5C%7D+& alt=&\left\{
\begin{array}{l}
k_1+k_2=-\dfrac{2b^2k}{m^2-b^2}\\
k_1k_2=\dfrac{b^2(m^2-a^2k^2)}{a^2(m^2-b^2)}
\end{array} \right\} & eeimg=&1&&&br&&p&由题设&/p&&blockquote&以&img src=&http://www.zhihu.com/equation?tex=A%2CB& alt=&A,B& eeimg=&1&&为直径的圆经过原点&/blockquote&可得&img src=&http://www.zhihu.com/equation?tex=k_1k_2%3D%5Cdfrac%7Bb%5E2%28m%5E2-a%5E2k%5E2%29%7D%7Ba%5E2%28m%5E2-b%5E2%29%7D%3D-1& alt=&k_1k_2=\dfrac{b^2(m^2-a^2k^2)}{a^2(m^2-b^2)}=-1& eeimg=&1&&&br&&p&所以&img src=&http://www.zhihu.com/equation?tex=b%5E2%28m%5E2-a%5E2k%5E2%29%3Da%5E2%28b%5E2-m%5E2%29%5CLeftrightarrow+%5Cdfrac%7B1%2Bk%5E2%7D%7Bm%5E2%7D%3D%5Cdfrac%7B1%7D%7Ba%5E2%7D%2B%5Cdfrac%7B1%7D%7Bb%5E2%7D& alt=&b^2(m^2-a^2k^2)=a^2(b^2-m^2)\Leftrightarrow \dfrac{1+k^2}{m^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}& eeimg=&1&&&/p&&p&而另一方面，原点到直线的距离平方：&/p&&img src=&http://www.zhihu.com/equation?tex=d%5E2%3D%5Cdfrac%7Bm%5E2%7D%7B1%2Bk%5E2%7D%3D%5Cdfrac%7Ba%5E2b%5E2%7D%7Ba%5E2%2Bb%5E2%7D& alt=&d^2=\dfrac{m^2}{1+k^2}=\dfrac{a^2b^2}{a^2+b^2}& eeimg=&1&&&br&&p&为定值。&/p&&br&&p&感谢&a href=&http://www.zhihu.com/people/306a02ed4ff& data-hash=&306a02ed4ff& class=&member_mention& data-hovercard=&p$b$306a02ed4ff&&@SoEasy&/a& 指出，虽然例题中的条件保证了使用到的点的横坐标&img src=&http://www.zhihu.com/equation?tex=x%5Cne0& alt=&x\ne0& eeimg=&1&&但是其他问题中是有可能出现&img src=&http://www.zhihu.com/equation?tex=x%3D0& alt=&x=0& eeimg=&1&&的情况的，这种情况单独拿出来讨论一下，说明一下结论正确即可。&/p&
引言我们知道，在常见的解析几何问题中，通常问题的解法都包含联立\rightarrow 伟达定理的步骤，即给定椭圆方程C：\begin{equation}
\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1
\end{equation} 和直线方程l：y=kx+m 通过联立获得\dfrac{b^2+a^2k^2}{a^2b^2}x^2+\…
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