Line 2aX by + 2 = 0 (a > 0, b > 0) bisector circle x ^ 2 + y + 2 + 2x-4y + 1 perimeter, find the minimum value of 1 / A + 1 / b

Line 2aX by + 2 = 0 (a > 0, b > 0) bisector circle x ^ 2 + y + 2 + 2x-4y + 1 perimeter, find the minimum value of 1 / A + 1 / b

The center of the circle x ^ 2 + y + 2 + 2x-4y + 1 = 0 is (- 1,2) according to the meaning of the title, a straight line passes through the center of the circle, so - 2a-2b + 2 = 0, that is, a + B = 1, so 1 / A + 1 / b = (a + b) / A + (a + b) / b = 1 + B / A + A / B + 1 = 2 + (B / A + A / b) ≥ 2 + 2 √ (B / a) * (A / b) = 2 + 2 = 4 if and only if B / a = A / B, that is, a = b = 1 / 2

If the line 2aX by + 2 = 0 (a, B ∈ R) always bisects the perimeter of x2 + Y2 + 2x-4y + 1 = 0, then the value range of a · B is ()
A. （-∞，14]B. （0，14）C. （0，14]D. （-∞，14）

∵ the line 2aX by + 2 = 0 (a, B ∈ R) always bisects the circumference of x2 + Y2 + 2x-4y + 1 = 0, and the center of the circle (- 1, 2) is on the line 2aX by + 2 = 0, we can get the solution of - 2a-2b + 2 = 0, we can get the solution of B = 1-A ∵ a · B = a (1-A) = - (A-12) 2 + 14 ≤ 14, if and only if a = 12, the equal sign holds, so the value range of a · B is (- ∞, 14], so we choose: a

Y = f (x) translate 2 units along the positive direction of X axis to get the curve C1, and the curve C1 is symmetric about y axis to get the curve C2

Y = f (x) moves 2 units along the positive direction of X axis, that is, 2 units to the right to get the curve C1
So C1 is y = f (X-2)
C1 curve C2 symmetric about y axis
So C2 is y = f [- (X-2)] = f (2-x)