椭圆x²/4+y²=1与圆(x-1)²+y²=r²有公共点,则半径r的
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椭圆x²/4+y²=1与圆(x-1)²+y²=r²有公共点,则半径r的最大值与最小值
椭圆与圆又公共点,说明下列方程组x,
x^2/4 + y^2 = 1 ---(1)
(x-1)^2 + y^2 = r^2 ---(2)
(2) - (1):
x^2 - 2x + 1 - x^2/4 = r^2 - 1
3/4 x^2 - 2x + (2 - r^2) = 0
判别式"b^2 - 4ac >= 0",所以:
(-2)^2 - 4*3/4*(2-r^2) >=0
4 - 3(2 - r^2) >= 0
4 - 6 + 3r^2 >= 0
r^2 >= 2/3
r >= √6/3
判别式"b^2 - 4ac" = 3r^2 - 2
x = (2 + - √(3r^2 - 2))/(2 * 3/4)
x = (4 + - 2√(3r^2 - 2))/3 ---(3)
(3)代入(1)得到关于y的一元二次方程,同理求判别式,令判别式>=0,
判别式计算过程从简,结果是:
7 - 3r^2 + 4√(3r^2 - 2) 或 7 - 3r^2 - 4√(3r^2 - 2)
7 - 3r^2 >= 4√(3r^2 - 2) ---(4)
或
7 - 3r^2 >= -4√(3r^2 - 2) ---(5)
解(4):
7 - 3r^2 >= 4√(3r^2 - 2)
(7 - 3r^2)^2 >= 16 (3r^2 - 2)
9r^4 -42r^2 + 49 >= 48r^2 + 32
9r^4 -90r^2 + 81 >= 0
r^4 -10r^2 + 9 >= 0
(r^2 -1)(r^2 -9) >= 0
r^2 >=9
r >= 3 ( 7 - 3r^2 为负值,舍去)
解(5):
7 - 3r^2 >= -4√(3r^2 - 2)
3r^2 - 7 <= 4√(3r^2 - 2)
(3r^2 - 7)^2 <= 16 (3r^2 - 2)
9r^4 -42r^2 + 49 <= 48r^2 + 32
9r^4 -90r^2 + 81 <= 0
r^4 -10r^2 + 9 <= 0
(r^2 -1)(r^2 -9) <= 0
r^2 <=9
r<=3
综上所述:
r >= √6/3 r最小值是√6/3
r <= 3 r最大值是 3
(完)
x^2/4 + y^2 = 1 ---(1)
(x-1)^2 + y^2 = r^2 ---(2)
(2) - (1):
x^2 - 2x + 1 - x^2/4 = r^2 - 1
3/4 x^2 - 2x + (2 - r^2) = 0
判别式"b^2 - 4ac >= 0",所以:
(-2)^2 - 4*3/4*(2-r^2) >=0
4 - 3(2 - r^2) >= 0
4 - 6 + 3r^2 >= 0
r^2 >= 2/3
r >= √6/3
判别式"b^2 - 4ac" = 3r^2 - 2
x = (2 + - √(3r^2 - 2))/(2 * 3/4)
x = (4 + - 2√(3r^2 - 2))/3 ---(3)
(3)代入(1)得到关于y的一元二次方程,同理求判别式,令判别式>=0,
判别式计算过程从简,结果是:
7 - 3r^2 + 4√(3r^2 - 2) 或 7 - 3r^2 - 4√(3r^2 - 2)
7 - 3r^2 >= 4√(3r^2 - 2) ---(4)
或
7 - 3r^2 >= -4√(3r^2 - 2) ---(5)
解(4):
7 - 3r^2 >= 4√(3r^2 - 2)
(7 - 3r^2)^2 >= 16 (3r^2 - 2)
9r^4 -42r^2 + 49 >= 48r^2 + 32
9r^4 -90r^2 + 81 >= 0
r^4 -10r^2 + 9 >= 0
(r^2 -1)(r^2 -9) >= 0
r^2 >=9
r >= 3 ( 7 - 3r^2 为负值,舍去)
解(5):
7 - 3r^2 >= -4√(3r^2 - 2)
3r^2 - 7 <= 4√(3r^2 - 2)
(3r^2 - 7)^2 <= 16 (3r^2 - 2)
9r^4 -42r^2 + 49 <= 48r^2 + 32
9r^4 -90r^2 + 81 <= 0
r^4 -10r^2 + 9 <= 0
(r^2 -1)(r^2 -9) <= 0
r^2 <=9
r<=3
综上所述:
r >= √6/3 r最小值是√6/3
r <= 3 r最大值是 3
(完)
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