Try to determine the range of real number a so that the positive root of equation x & # 178; - ax + A & # 178; - 4 = 0 has and only has one?

Try to determine the range of real number a so that the positive root of equation x & # 178; - ax + A & # 178; - 4 = 0 has and only has one?

Y = x & # 178; - ax + A & # 178; - 4 is a parabola with opening upward, and there is only one positive root
There are two situations
(1) One positive root, one negative root
x=0 x²-ax+a²-4=a^2-4

If x + y-radical (x + y) + 3K = 0 represents only one straight line, then the value range of K is?

∵ x + Y - √ (x + y) + 3K = 0 let √ x + y = t, then T-T + 3K = = 0 ∵ denotes a straight line, ∵ x + y is the only value, that is, t is the only value, that is, the equation T-T + 3K = 0 has a unique solution ∵ Δ= 1-12k = 0 ∵ k = 1 / 12

The trajectory equation of the center of a circle which is circumscribed with the circle x2 + y2-4y = 0 and tangent to the x-axis is ()
A. Y2 = 8xb. Y2 = 8x (x > 0) and y = 0C. X2 = 8y (Y > 0) d. x2 = 8y (Y > 0) and x = 0 (y < 0)

According to the meaning of the problem, let the center m coordinate of the circle be m (x, y), ∵ the circle and the circle C: x2 + y2-4y = 0, that is, X2 + (Y-2) 2 = 4 circumtangent, and also tangent to the X axis, | MC | = | y | + 2 | x2 + (Y − 2) 2 = 2 + | y |, | x2 + y2-4y + 4 = 4 + 4 | y | + Y2, | x2 = 4Y + 4 | y |, when y > 0, X2 = 8y; when y < 0

The trajectory equation of the center of a circle which is circumscribed with the circle x2 + y2-4y = 0 and tangent to the x-axis is ()
A. Y2 = 8xb. Y2 = 8x (x > 0) and y = 0C. X2 = 8y (Y > 0) d. x2 = 8y (Y > 0) and x = 0 (y < 0)

According to the meaning of the problem, let the center m coordinate of the circle be m (x, y), ∵ the circle and the circle C: x2 + y2-4y = 0, that is, X2 + (Y-2) 2 = 4 circumtangent, and also tangent to the X axis, | MC | = | y | + 2 | x2 + (Y − 2) 2 = 2 + | y |, | x2 + y2-4y + 4 = 4 + 4 | y | + Y2, | x2 = 4Y + 4 | y |, when y > 0, X2 = 8y; when y < 0