If the line 2ax-by + 2 = 0 (where a and B are positive real numbers) passes through the center of the circle C: x2 + Y2 + 2x-4y + 1 = 0, then the minimum value of 4A + 1b is______ ．

If the line 2ax-by + 2 = 0 (where a and B are positive real numbers) passes through the center of the circle C: x2 + Y2 + 2x-4y + 1 = 0, then the minimum value of 4A + 1b is______ ．

The center (- 1,2) of circle x2 + Y2 + 2x-4y + L = 0 is on line 2aX by + 2 = 0, so - 2a-2b + 2 = 0, that is, 1 = a + B is substituted by 4A + 1b, and (4a + 1b) (a + b) = 5 + 4BA + ab ≥ 9 (a > 0, b > 0, if and only if a = 2B, take the equal sign), so the answer is: 9

Let a (- 3,5) and B (2,15) find a point P on the line L: 3x-4y + 4 = 0, so that | PA | + | Pb | is the minimum
Let points a (- 3,5) and B (2,15), find a point P on the line L: 3x-4y + 4 = 0, make | PA | + | Pb | the minimum, and find the minimum

P (8 / 3,3) (| PA | + | Pb |) min = 5 √ 17. Because the distance between two points is the shortest, we only need to make a '(x0, Y0) symmetric point a' (or B point) about the line L, and then connect a'B with the intersection point of the line L to obtain P. first, we need to find a '(x0, Y0) 3 (x0-3) / 2 - 4 (Y0 + 5) / 2 + 4 = 0 (1) formula (y0-5) / (x0 + 3) = -