If x ^ 2 + y ^ 2-8x-10y + 41 = 0, find the value of X / y + Y / X Please write that I am a junior one, please use the knowledge of junior one to answer

If x ^ 2 + y ^ 2-8x-10y + 41 = 0, find the value of X / y + Y / X Please write that I am a junior one, please use the knowledge of junior one to answer

x^2+y^2-8x-10y+41=0
x^2-8x+16+y^2-10y+25=0
(x-4)^2+(y-5)^2=0
So x = 4, y = 5
x/y+y/x＝4/5+5/4=0.8+1.25=2.05

It is known that O is the coordinate origin, vector ob = (2cos square x.1), vector ob = (1, √ 3sin2x + a) (x belongs to R, a belongs to R, a is a constant)
Given that O is the origin of the coordinate, the vector ob = (the square X of 2cos, 1), the vector ob = (1, √ 3sin2x + a) (x belongs to R, a belongs to R, a is a constant), and y = the vector OA times the vector ob
(1) The relation of function x (x) is obtained;
(2) The maximum value of NOF (x) is 2, and the value of a is obtained;
(3) By using the conclusion of (2), the function f (x) on a closed interval of one period is drawn by "five point method", and its monotone interval is pointed out

(1)y = OA.OB= (2(cosx)^2,1).(1,√3sin2x+a）= 2(cosx)^2 + √3sin2x+a(2)y' = -4cosxsinx + 2√3cos2x =0sin2x-√3cos2x=0tan2x = √3x = π/6y''(π/6) 3+a =2a = -1(3)sorry,I cannot draw this graph

Why can the plane rectangular coordinate system of unit vector be expressed as the square of abscissa plus the square of ordinate equal to one?

The hypotenuse is the line segment where the unit vector is located. Obviously, X and Y in its (x, y) are its right angles. Because it is a unit vector, module = 1, that is to say, X & sup2; + Y & sup2; = 1 & sup2; = 1
This is an algorithm for the module of a vector

Does the unit vector have to be equal to 1
A vector of module 1 must be a unit vector
The unit vector must be equal to 1
Which of these two sentences is right and which is wrong? What are the conditions before and after

A vector of module 1 must be a unit vector
Because the unit vector can only be 1, but the module 1 can be - 1 or 1