Let circle: x ^ 2 + y ^ 2 - (2a ^ 2-4) x-4a ^ 2Y + 5A ^ 4-4 = 0, find the value range of real number a and the trajectory equation of center C

Let circle: x ^ 2 + y ^ 2 - (2a ^ 2-4) x-4a ^ 2Y + 5A ^ 4-4 = 0, find the value range of real number a and the trajectory equation of center C

[x-(a^2-2)]^2+(y-2a^2)=4-5a^4+(a^2-2)^2+4a^4
=8-4a^2≥0
a^2≤2
-√2≤a≤√2
x=a^2-2
y=2a^2
x=y/2-2
y=2x+4

4a^3-3a^2

1. A ^ 2 (4a-3) = 0 and a = 0 does not satisfy the inequality (left = 0)
2. When a > 0
4a-3

If the three sides of a triangle are 3a, 4A and 14, then the value range of a is the same

4a-3a14
two

The equation x2 + Y2 + ax + 2ay + 2A2 + A-1 = 0 represents a circle, then the value range of a is ()
A. a＜-2B. -23＜a＜0C. -2＜a＜0D. -2＜a＜23

The equation x2 + Y2 + ax + 2ay + 2A2 + A-1 = 0 indicates that the circle 〈 A2 + 4a2-4 (2A2 + A-1) ＞ 0 〉 3a2 + 4a-4 ＜ 0, 〈 (a + 2) (3a-2) ＜ 0, 〈 − 2 ＜ a ＜ 23, so D is selected