It is known that real numbers a, x, y satisfy the trajectory equation of a ^ + 2A + 2XY + (a + X-Y) I = 0, (x, y) Come on. It's urgent. Thank you

It is known that real numbers a, x, y satisfy the trajectory equation of a ^ + 2A + 2XY + (a + X-Y) I = 0, (x, y) Come on. It's urgent. Thank you

The coefficient of imaginary number I is 0
So a + X-Y = 0, a = y-x
The binary equation x ^ 2 + y ^ 2 + 2y-2x = 0 for X and Y is obtained
This is a circular equation, written in the standard form, which is (x-1) ^ 2 + (Y-1) ^ 2 = 2
I do not know if there are doubts, welcome to ask ~!

1. The image of quadratic function passes through the point (- 1,0) (3,0), and the maximum value is 3
Passing point (- 1,0) (3,0)
That is to say, the abscissa of the intersection point with X axis is - 1 and 3
So - 1 and 3 are solutions of the equation y = ax ^ 2 + BX + C = 0
So y = a [x - (- 1)] (x-3) = a (x ^ 2-2x-3)
=a[(x-1)^2-4]
=a(x-1)^2-4a
The maximum value is 3
So - 4A = 3
a=-3/4
y=-(3/4)(x^2-2x-3)
=-3x^2/4+3x/2+4/9
Solution: the parabola passes through the point (- 1,0), (3,0),
So the axis of symmetry is x = 1
The vertex is (1,3)
Let the analytic formula be y = a (x-1) & sup2; + 3
If (3,0) is brought into the analytic expression, we can get
0=4a+3
a=-3/4
So the parabola analytic formula is y = - 3 / 4 (x-1) & sup2; + 3
Which of these two is right?

Both methods are correct, but the result of the first method is a little wrong (the last step). Very good, I can think of two different methods, come on

When x = - 1, the function has a maximum value of 2,
Write out the calculation process

Let y = a (x + 1) ^ 2 + 2,
Substituting (- 2, - 1) into,
a(-2-1)^2+2=-1,
a=-1/3
Analytical formula: y = - x ^ 2 / 3-2x / 3 + 5 / 3