已知椭圆C:X^2/(a^2)+y^2/(b^2)=1(a>b>0)离心率1/2,以原点为圆点,椭圆的短轴为半径圆与直线x-y+根号6相切,设p(4,0),A,B椭圆C上关于X轴对称的任意两个不同的点,连接PB交椭圆C于另一点E,证明AE与X轴相交
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![已知椭圆C:X^2/(a^2)+y^2/(b^2)=1(a>b>0)离心率1/2,以原点为圆点,椭圆的短轴为半径圆与直线x-y+根号6相切,设p(4,0),A,B椭圆C上关于X轴对称的任意两个不同的点,连接PB交椭圆C于另一点E,证明AE与X轴相交](/uploads/image/z/3706718-14-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%A4%AD%E5%9C%86C%3AX%5E2%2F%28a%5E2%29%2By%5E2%2F%28b%5E2%29%3D1%28a%3Eb%3E0%29%E7%A6%BB%E5%BF%83%E7%8E%871%2F2%2C%E4%BB%A5%E5%8E%9F%E7%82%B9%E4%B8%BA%E5%9C%86%E7%82%B9%2C%E6%A4%AD%E5%9C%86%E7%9A%84%E7%9F%AD%E8%BD%B4%E4%B8%BA%E5%8D%8A%E5%BE%84%E5%9C%86%E4%B8%8E%E7%9B%B4%E7%BA%BFx-y%2B%E6%A0%B9%E5%8F%B76%E7%9B%B8%E5%88%87%2C%E8%AE%BEp%284%2C0%29%2CA%2CB%E6%A4%AD%E5%9C%86C%E4%B8%8A%E5%85%B3%E4%BA%8EX%E8%BD%B4%E5%AF%B9%E7%A7%B0%E7%9A%84%E4%BB%BB%E6%84%8F%E4%B8%A4%E4%B8%AA%E4%B8%8D%E5%90%8C%E7%9A%84%E7%82%B9%2C%E8%BF%9E%E6%8E%A5PB%E4%BA%A4%E6%A4%AD%E5%9C%86C%E4%BA%8E%E5%8F%A6%E4%B8%80%E7%82%B9E%2C%E8%AF%81%E6%98%8EAE%E4%B8%8EX%E8%BD%B4%E7%9B%B8%E4%BA%A4)
已知椭圆C:X^2/(a^2)+y^2/(b^2)=1(a>b>0)离心率1/2,以原点为圆点,椭圆的短轴为半径圆与直线x-y+根号6相切,设p(4,0),A,B椭圆C上关于X轴对称的任意两个不同的点,连接PB交椭圆C于另一点E,证明AE与X轴相交
已知椭圆C:X^2/(a^2)+y^2/(b^2)=1(a>b>0)离心率1/2,以原点为圆点,椭圆的短轴为半径圆与直线x-y+根号6相切,设p(4,0),A,B椭圆C上关于X轴对称的任意两个不同的点,连接PB交椭圆C于另一点E,证明AE与X轴相交于定点Q
已知椭圆C:X^2/(a^2)+y^2/(b^2)=1(a>b>0)离心率1/2,以原点为圆点,椭圆的短轴为半径圆与直线x-y+根号6相切,设p(4,0),A,B椭圆C上关于X轴对称的任意两个不同的点,连接PB交椭圆C于另一点E,证明AE与X轴相交
设M(x1,y1),N(x2,y2)
联立直线椭圆,得:
(1+2k²)x² - 4k²x+2k²-4=0
x1+x2=4k²/(1+2k²),x1x2=(2k²-4)/(1+2k²)
|MN|=√[(x1-x2)²+(y1-y2)²]
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设M(x1,y1),N(x2,y2)
联立直线椭圆,得:
(1+2k²)x² - 4k²x+2k²-4=0
x1+x2=4k²/(1+2k²),x1x2=(2k²-4)/(1+2k²)
|MN|=√[(x1-x2)²+(y1-y2)²]
=√{ (x1-x2)² + [k(x1-1) - k(x2-1)]² }
=√[(x1-x2)² + k²(x1-x2)²]
=√[(1+k²)(x1-x2)²]
=√{ (1+k²)[(x1+x2)² - 4x1x2]
=√{ (1+k²)[16k^4/(1+2k²)² - 4(2k²-4)/(1+2k²) ] }
=√[(1+k²)(24k²+16)/(1+2k²)² ]
A点到直线距离为
h=|k|/√(1+k²)
∴S=(1/2)·h·|MN|
=(1/2)·[|k|/√(1+k²)] ·√[(1+k²)(24k²+16)/(1+2k²)² ]
=(1/2)·|k|·√[(24k²+16)/(1+2k²)²]
=√10/3
即:|k|·√[(24k²+16)/(1+2k²)²] = 2√10/3
两边平方,得:(24k^4 + 16k²)/(1+2k²)² = 40/9
即:7k^4 - 2k² - 5=0
解得:k²=1或-5/7 (舍去)
∴k²=1
∴k=±1望采纳
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