这道题的第三小题怎么做
2个回答
展开全部
(1)
A(1,4)
过纤指橘点逗嫌B(-3,1) 的直线方程l:
y-1=m(x+3)
mx -y +3m+1 =0
A与l的距离=d
|m -4 +3m+1|/√(m^2+1) =d
|4m-3|/√(m^2+1) =d
(4m-3)^2 =d^2.(m^2+1)
(16- d^2)m^2 -24m +(9-d^2) =0
△≥0
(-24)^2 -4(16- d^2)(9-d^2) ≥0
144-(16- d^2)(9-d^2) ≥毁团0
144 -( 144 -25d^2 +d^4) ≥0
d^4 -25d^2 ≤0
d^2-25≤0
-5≤d≤5
(2)
max d =5
|4m-3|/√(m^2+1) =5
(4m-3)^2 =25(m^2+1)
9m^2 +24m+16 =0
(3m+4)^2=0
m=-4/3
直线方程l:
y-1=m(x+3)
y-1 = -(4/3)(x+3)
-3y+3 = 4x+12
4x+3y+9=0
(3)
d=4
|4m-3|/√(m^2+1) =4
(4m-3)^2 =16(m^2+1)
24m+7=0
m=-7/24
直线方程l:
y-1=m(x+3)
=-(7/24)(x+3)
-24y+24 =7x+21
7x+24y-3=0
A(1,4)
过纤指橘点逗嫌B(-3,1) 的直线方程l:
y-1=m(x+3)
mx -y +3m+1 =0
A与l的距离=d
|m -4 +3m+1|/√(m^2+1) =d
|4m-3|/√(m^2+1) =d
(4m-3)^2 =d^2.(m^2+1)
(16- d^2)m^2 -24m +(9-d^2) =0
△≥0
(-24)^2 -4(16- d^2)(9-d^2) ≥0
144-(16- d^2)(9-d^2) ≥毁团0
144 -( 144 -25d^2 +d^4) ≥0
d^4 -25d^2 ≤0
d^2-25≤0
-5≤d≤5
(2)
max d =5
|4m-3|/√(m^2+1) =5
(4m-3)^2 =25(m^2+1)
9m^2 +24m+16 =0
(3m+4)^2=0
m=-4/3
直线方程l:
y-1=m(x+3)
y-1 = -(4/3)(x+3)
-3y+3 = 4x+12
4x+3y+9=0
(3)
d=4
|4m-3|/√(m^2+1) =4
(4m-3)^2 =16(m^2+1)
24m+7=0
m=-7/24
直线方程l:
y-1=m(x+3)
=-(7/24)(x+3)
-24y+24 =7x+21
7x+24y-3=0
本回答被网友采纳
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