The distance from the moving point to the point (1,0) is equal to the distance from the moving point to the straight line y + 2 = 0

The distance from the moving point to the point (1,0) is equal to the distance from the moving point to the straight line y + 2 = 0

Let the moving point be (x, y)
Then the distance from it to (1,0) is √ ((x-1) ^ 2 + y ^ 2)
The distance to the line y + 2 = 0 is y + 2
The two are equal
((x-1)^2+y^2=(y+2)^2
The trajectory equation is obtained
x^2-2x-3=4y

Given that the distance between point m and straight line x + 1 = 0 is equal to the distance between point m and straight line Y-1 = 0, then the trajectory equation of point m?

Set point m (x, y)
The distance between point m and line x + 1 = 0 is equal to the distance between point m and line Y-1 = 0
∴│x+1│=│y-1│
∴（x+1）^2=(y-1)^2
∴（x+1+y-1）（x+1－y＋1）=0
∴（x+y）（x-y+2）=0
The trajectory equation of point m is x + y = 0 or X-Y + 2 = 0

Given that the domain of the function f (x) = 1 / LG (2 ^ x + 4 * 2 ^ - x-a) is r, the range of the real number a is obtained

First, 2 ^ x + 4 * (2 ^ - x) - a > 0, that is, A0 or T ^ 2 - (a + 1) t + 40, then a + 1

Big question We know that the function y = ax to the third power + BX to the second power, when x = 1, there is a maximum of 3 Finding the value of a and B

Y = ax & sup3; + BX & sup2;, y '= 3ax & sup2; + 2bx,
The derivative at the extreme point is 0,3a × 1 & sup2; + 2B × 1 = 0
When x = 1, there is a maximum of 3, so a + B = 3
The solution is a = - 6, B = 9