2个回答
2013-01-20
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将y = x + m代入4x² + y² = 1中
4x² + (x + m)² = 1
5x² + 2mx + (m² - 1) = 0
有公共点即Δ ≥ 0
[2m]² - 4[5][m² - 1] ≥ 0
- 4(4m² - 5) ≥ 0
4m² - 5 ≤ 0
m² ≤ 5/4
- √5/2 ≤ m ≤ √5/2
当m = 1时,y = x + 1,同样代入
4x² + (x + 1)² = 1
5x² + 2x = 0
x = - 2/5 或 x = 0
y = 3/5 或 y = 1
弦长 = √[(- 2/5 - 0)² + (3/5 - 1)²] = (2√2)/5
4x² + (x + m)² = 1
5x² + 2mx + (m² - 1) = 0
有公共点即Δ ≥ 0
[2m]² - 4[5][m² - 1] ≥ 0
- 4(4m² - 5) ≥ 0
4m² - 5 ≤ 0
m² ≤ 5/4
- √5/2 ≤ m ≤ √5/2
当m = 1时,y = x + 1,同样代入
4x² + (x + 1)² = 1
5x² + 2x = 0
x = - 2/5 或 x = 0
y = 3/5 或 y = 1
弦长 = √[(- 2/5 - 0)² + (3/5 - 1)²] = (2√2)/5
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