数学题目 求解答 20
展开全部
联立直线与椭圆方程有:4x²+(x+b)²=1
==> 4x²+x²+2bx+b²-1=0
==> 5x²+2bx+(b²-1)=0
==> x1+x2=-2b/5,x1x2=(b²-1)/5
所以,(x1-x2)²=(x1+x2)²-4x1x2=(4b²/25)-4(b²-1)/5=(20-16b²)/25
且,(y1-y2)²=[(x1+b)-(x2+b)]²=(x1-x2)²
那么,截得的弦长=√[(x1-x2)²+(y1-y2)²]=√[2(x1-x2)²]=2/√5
==> (x1-x2)²=2/5
==> (20-16b²)/25=2/5
==> 20-16b²=10
==> b=±√10/4
==> 4x²+x²+2bx+b²-1=0
==> 5x²+2bx+(b²-1)=0
==> x1+x2=-2b/5,x1x2=(b²-1)/5
所以,(x1-x2)²=(x1+x2)²-4x1x2=(4b²/25)-4(b²-1)/5=(20-16b²)/25
且,(y1-y2)²=[(x1+b)-(x2+b)]²=(x1-x2)²
那么,截得的弦长=√[(x1-x2)²+(y1-y2)²]=√[2(x1-x2)²]=2/√5
==> (x1-x2)²=2/5
==> (20-16b²)/25=2/5
==> 20-16b²=10
==> b=±√10/4
展开全部
y=x+b (1)
4x^2+y^2 =1 (2)
sub (1) into (2)
4x^2+(x+b)^2 -1 =0
5x^2+2bx + b^2-1 =0
x1+x2 = -2b/5
x1.x2 = (b^2-1)/5
(x1-x2)^2 =(x1+x2)^2 - 4x1.x2
= 4b^2/25 - (4/5)(b^2-1)
= 4/5 - (16/25)b^2 (3)
sub (1) into (2)
4(y-b)^2+y^2 =1
5y^2-8by +4b^2 -1=0
y1+y2= 8b/5
y1.y2 = (4b^2 -1)/5
(y1-y2)^2= (y1+y2)^2 -4y1.y2
= (64/25)b^2 - (4/5)(4b^2 -1)
=4/5 - (16/25)b^2 (4)
截距=2/√5
(x1-x2)^2 +(y1-y2)^2 = 4/5
4/5 - (16/25)b^2 +4/5 - (16/25)b^2 =4/5
(32/25)b^2 = 4/5
b^2 = 5/8
b= (1/4)√10 or -(1/4)√10
4x^2+y^2 =1 (2)
sub (1) into (2)
4x^2+(x+b)^2 -1 =0
5x^2+2bx + b^2-1 =0
x1+x2 = -2b/5
x1.x2 = (b^2-1)/5
(x1-x2)^2 =(x1+x2)^2 - 4x1.x2
= 4b^2/25 - (4/5)(b^2-1)
= 4/5 - (16/25)b^2 (3)
sub (1) into (2)
4(y-b)^2+y^2 =1
5y^2-8by +4b^2 -1=0
y1+y2= 8b/5
y1.y2 = (4b^2 -1)/5
(y1-y2)^2= (y1+y2)^2 -4y1.y2
= (64/25)b^2 - (4/5)(4b^2 -1)
=4/5 - (16/25)b^2 (4)
截距=2/√5
(x1-x2)^2 +(y1-y2)^2 = 4/5
4/5 - (16/25)b^2 +4/5 - (16/25)b^2 =4/5
(32/25)b^2 = 4/5
b^2 = 5/8
b= (1/4)√10 or -(1/4)√10
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询