圆锥曲线与直线方程 圆锥曲线与方程
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\u9898\u76ee\u8bf4\u76f4\u7ebfl:2x+y+2=0\u5173\u4e8e\u539f\u70b9\u5bf9\u79f0\u7684\u76f4\u7ebf\u4e3al'
l'\uff08AB\uff09-2x-y+2=0
y=-2x+2
\u82e5l'\u4e0e\u692d\u5706x^2+y^2/4=1\u7684\u4ea4\u70b9\u4e3aA,B
a=2,b=1
\u6b64\u76f4\u7ebf\u7ecf\u8fc7\u692d\u5706\u4e0a\u9876\u70b9A(0,2)\uff0c\u53f3\u9876\u70b9B(1,0)
AB=\u221a5
S\u25b3PAB=1/2hAB=1/2
h=\u221a5/5
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\u8bbe\u5e73\u884c\u76f4\u7ebf2x+y+c=0
| c-\uff08-2\uff09 |/\u221a5=\u221a5/5\uff08\u5e73\u884c\u7ebf\u8ddd\u79bb\u516c\u5f0f\uff09
c=-1
\u5373\u76f4\u7ebf2x+y-1=0
y=-2x+1
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\u5373p\u4e2a\u6570\u4e3a2
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解：联立方程组y=kx+b和x^2/4+y^2=/1
得（4k^2+1）x^2+8kbx+4b^2…………………………（你应该会化吧- -）
判别式=16（4k^2-b^2+1）
AB=根号下（1+k^2）*|x1-x2|=2………………………………①
又∵O到AB得距离d=|b|/根号下(1+k^2)=2S/|AB|=1∴b^2=k^2+1………………………………②
把②代入①，得4k^4-4k^2+1=0
解得k^2=1/2，b^2=3/2
经检验判别式＞0（这部必须写！！！！！！！！！！！！）
∴直线方程是…………………………你应该会了吧
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