Question

已知曲线y^2=ax与其关于点(1,1)对称的曲线有两个不同的交点

已知曲线y^2=ax与关于点(1,1)对称的曲线有两个不同的交点.如果过这两个交电的直线的倾斜角为45度,求A的值

Answer 1

设y² = ax上有任一点A(a, b), 其关于(1, 1)的对称点为A'(x, y):
(a + x)/2 = 1, a = 2 - x
(b + y)/2 = 2, b = 2 - y
即y² = ax变为: (2 - y)² = a(2 - x) = 4a - 2ax = 4a - y²
y² - 2y + 2 - a = 0
y = 1 ± √(a - 1)
x = 1 ± (2/a)√(a - 1)
两个不同的交点: (1 + (2/a)√(a - 1), 1 + √(a - 1)), (1 - (2/a)√(a - 1), 1 - √(a - 1))
过这两个交电的直线的斜率=tan45˚ = 1
= [1 + √(a - 1) - 1 + √(a - 1)]/[1 + (2/a)√(a - 1) - 1 + (2/2)√(a - 1)]
= a/2
a = 2