求解一道圆锥曲线题目...
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求解一道圆锥曲线题目...
如图.
如图.
![求解一道圆锥曲线题目...](/uploads/image/z/17548203-3-3.jpg?t=%E6%B1%82%E8%A7%A3%E4%B8%80%E9%81%93%E5%9C%86%E9%94%A5%E6%9B%B2%E7%BA%BF%E9%A2%98%E7%9B%AE...)
2.代y=kx+1入椭圆方程4x^2+y^2-4=0中,得(4+k^2)x^2+2kx-3=0
y1y2=k^2x1x2+k(x1+x2)+1
x1x2+y1y2=(1+k^2)x1x2+k(x1+x2)+1
x1x2=-3/(4+k^2) x1+x2=-2k/(4+k^2)
x1x2+y1y2=(1+k^2)x1x2+k(x1+x2)+1
=-3*(1+k^2)/(4+k^2)-2k^2*/(4+k^2)+1=0
4k^2=1 k=±1/2
|AB|=根号(1+k^2)*根号[(x1+x2)^2-4x1x2]
=根号(5/4)*根号[(4/17)^2+48/17]=【4*根号65】/17
y1y2=k^2x1x2+k(x1+x2)+1
x1x2+y1y2=(1+k^2)x1x2+k(x1+x2)+1
x1x2=-3/(4+k^2) x1+x2=-2k/(4+k^2)
x1x2+y1y2=(1+k^2)x1x2+k(x1+x2)+1
=-3*(1+k^2)/(4+k^2)-2k^2*/(4+k^2)+1=0
4k^2=1 k=±1/2
|AB|=根号(1+k^2)*根号[(x1+x2)^2-4x1x2]
=根号(5/4)*根号[(4/17)^2+48/17]=【4*根号65】/17