双曲线4x^2-y^2=1,（1）求以该双曲线的实轴和虚轴分别为短轴和长轴的椭圆方程．（2）证明当与直线y=x-3平行的直线被椭圆截取的弦最长时,则此直线被双曲线截取的弦为最短．求此直线方程及

Question

双曲线4x^2-y^2=1,（1）求以该双曲线的实轴和虚轴分别为短轴和长轴的椭圆方程．（2）证明当与直线y=x-3平行的直线被椭圆截取的弦最长时,则此直线被双曲线截取的弦为最短．求此直线方程及此时被椭圆截取的最长弦长和被双曲线截取的最短弦长

Answer 1

1、x^2/(1/4) - y^2 = 1, 实轴和虚轴分别为1/2, 1
∴椭圆的短轴和长轴分别为1/2, 1 , 且焦点在y轴
∴x^2/(1/4) + y^2 = 1 ,即：4x^2 + y^2 = 1
2、设直线方程为y = x + b ,
截椭圆、双曲线的弦分别为AB、CD,
A(x1,y1)、B(x2,y2)、C(x3,y3)、D(x4,y4)
联立椭圆、直线,得：
5x^2 +2bx + b^2-1=0
△ = 4b^2 - 20(b^2-1) ＞0 , 即：-√5/2 ＜b ＜√5/2
x1+x2 = -2b/5, x1x2 = (b^2-1)/5
|AB|^2= (x1-x2)^2 + (y1-y2)^2
=(x1-x2)^2 + (y1-y2)^2
=(x1-x2)^2 + [(x1+b)-(x2+b)]^2
=2(x1-x2)^2
=2[(x1+x2)^2 - 4x1x2]
=2(2b/5)^2 - 8(b^2-1)/5
=(40-32b^2)/25
≤40/25 = 8/5
当且仅当b = 0时取等号.
联立直线、双曲线,得：
3x^2 - 2bx -(b^2 + 1) =0
△ = 4b^2 + 12(b^2 + 1) = 16b^2 + 12 ＞ 0 恒成立
x3+x4 = 2b/3 , x3x4 = -(b^2 + 1)/3
|CD|^2 = (x3-x4)^2 + (y3-y4)^2
=(x3-x4)^2 + [(x3+b)-(x4+b)]^2
=2(x3-x4)^2
=2[(x3+x4)^2 - 4x3x4]
=2(2b/3)^2 + 8(b^2+1)/3
=(32b^2 + 8)/9
≥ 8/9
当且仅当b = 0时取等号.
∴当 b = 0时,直线被椭圆截取的弦最长为(2/5)√10,被双曲线截取的弦最短(2/3)√2
此时直线方程为y = x