九州算术数学论坛 » 高中数学 » 圆锥曲线的切线怎么求？

IMath 发表于 2007-12-25 16:43

圆锥曲线的切线怎么求？

圆锥曲线的切线怎么求？<br>

xiaxiao 发表于 2007-12-25 16:43

<img src="http://www.oxtoo.com/tmp/0712/055413_8zCOqjf1Dnp4DmPrJkdH.GIF" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br>

fanyun 发表于 2007-12-25 16:43

根据条件设出只带一个参数的直线，与圆锥曲线方程联立，消去一个变量后得一元二次方程，Δ=0求得参数<br>

Chelly 发表于 2007-12-25 16:43

好像不是很清楚，<br />FB+FA=AA'+BB'=2OO'=4<br />不过与x轴的两个交点不能要<br>

85463295 发表于 2007-12-25 16:43

学过导数的话，也可以求导求斜率<br>

IMath 发表于 2007-12-25 16:43

第二题字母标错了吧，|AN|=l2?<br /><br />设圆心 <img src="http://www.imathas.com/cgi-bin/mimetex.cgi? C(2pt,2pt^2)" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br />圆方程 <img src="http://www.imathas.com/cgi-bin/mimetex.cgi? (x-2pt)^2+(y-2pt^2)^2=(2pt)^2+(p-2pt^2)^2" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br />即 <img src="http://www.imathas.com/cgi-bin/mimetex.cgi? (x-2pt)^2+y^2-4pt^2y=p^2" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br />令 <img src="http://www.imathas.com/cgi-bin/mimetex.cgi? y=0" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /> 得 <img src="http://www.imathas.com/cgi-bin/mimetex.cgi? (x-2pt)^2=p^2" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br />即 <img src="http://www.imathas.com/cgi-bin/mimetex.cgi? x_1=2pt-p;\quad x_2=2pt+p" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br />即 <img src="http://www.imathas.com/cgi-bin/mimetex.cgi? l_1=\sqrt{p^2+x_1^2}=p\sqrt{4t^2-4t+2};\quad l_2=p\sqrt{4t^2+4t+2}" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br /><br /><img src="http://www.imathas.com/cgi-bin/mimetex.cgi? \frac{l_1}{l_2}+\frac{l_2}{l_1}=\sqrt{\frac{2t^2-2t+1}{2t^2+2t+1}}+\sqrt{\frac{2t^2+2t+1}{2t^2-2t+1}}" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br /><br />最小值是2，最大值是<img src="http://www.imathas.com/cgi-bin/mimetex.cgi? 2\sqrt{2}" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br>

ZhiJie 发表于 2007-12-25 16:43

利用对称性也是求圆锥曲线切线的通法<br>

ZhiJie 发表于 2007-12-25 16:43

对，标错了，是|AN|<br>

Okppt 发表于 2007-12-25 16:43

没用过，能举例一下看看么。。。。。<br>

85463295 发表于 2007-12-25 16:43

我是先证明MN为定值，等于2p，用勾股定理求的，然后再用不等式（或耐克函数）做的<br>

nihoo 发表于 2007-12-25 16:43

提示一下,若两圆相切是两圆相减表示什么<br>

Chelly 发表于 2007-12-25 16:43

求l1/l2的范围，再求与倒数加的范围<br>

IMath 发表于 2007-12-25 16:43

哦。。。。。学习了<br>

xiaxiao 发表于 2007-12-25 16:43

<br><br>QUOTE:原帖由 oldshanmao 于 2007-12-23 02:12 发表 <a href="http://bbs.pep.com.cn/redirect.php?goto=findpost&pid=3541638&ptid=346863" target="_blank"><img src="http://bbs.pep.com.cn/images/common/back.gif" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /></a><br />第二题字母标错了吧，|AN|=l2?<br /><br />设圆心 <a href="http://www.imathas.com/cgi-bin/mimetex.cgi?" target="_blank">http://www.imathas.com/cgi-bin/mimetex.cgi?</a> C(2pt,2pt^2)<br />圆方程 <a href="http://www.imathas.com/cgi-bin/mimetex.cgi?" target="_blank">http://www.imathas.com/cgi-bin/mimetex.cgi?</a> (x-2pt)^2+(y-2pt^2)^2=(2pt)^2+(p-2pt^2)^2<br />即 h ... <br>可求得MN=2p<br /><img src="http://www.imathas.com/cgi-bin/mimetex.cgi? \frac{l_{1}}{l_{2}}+\frac{l_{2}}{l_{1}}=\frac{l_{1}^{2}+l_{2}^{2}}{l_{1}*l_{2}}=\frac{l_{1}^{2}+l_{2}^{2}-MN^{2}}{l_{1}*l_{2}}+\frac{MN^{2}}{l_{1}*l_{2}}" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br />因为<img src="http://www.imathas.com/cgi-bin/mimetex.cgi? \frac{l_{1}^{2}+l_{2}^{2}-MN^{2}}{l_{1}*l_{2}}=2*cos\theta" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br />且<img src="http://www.imathas.com/cgi-bin/mimetex.cgi? S=\frac{1}{2}*(l_{1}*l_{2})*sin\theta=\frac{1}{2}*OA*MN=\frac{1}{4}*MN^{2}\Rightarrow\frac{MN^{2}}{l_{1}*l_{2}}=2*sin\theta" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br />所以<img src="http://www.imathas.com/cgi-bin/mimetex.cgi? \frac{l_{1}}{l_{2}}+\frac{l_{2}}{l_{1}}=2*cos\theta+2*sin\theta" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br />其中<img src="http://www.imathas.com/cgi-bin/mimetex.cgi? \theta=MAN" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /><br />接下来就简单了<br /><br />[ 本帖最后由 zchpeter 于 2007-12-23 13:16 编辑 ]<br>

xiaxiao 发表于 2007-12-25 16:43

过切点的公切线方程<br>

Chengqin1987 发表于 2007-12-25 16:43

怎么打上符号的？<br>

85463295 发表于 2007-12-25 16:43

有一个统一的表达式。<br>

chfxm 发表于 2008-2-18 22:26

回复 undefined 的帖子

哪来什么第二题，晕:@

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