Given that i, j is the unit vector in the positive direction of x, y axis, let vector a=(x-root number 3) i+yj, vector b=(x+root number 3) i+yj, and satisfy direction |a||b|=4 Finding the Trajectory Equation of Point P (x, y) Given that i, j are unit vectors in the positive direction of x and y axes, let vector a=(x-root number 3) i+yj, vector b=(x+root number 3) i+yj, and satisfy direction |a||b|=4 Finding the Trajectory Equation of Point P (x, y)

Given that i, j is the unit vector in the positive direction of x, y axis, let vector a=(x-root number 3) i+yj, vector b=(x+root number 3) i+yj, and satisfy direction |a||b|=4 Finding the Trajectory Equation of Point P (x, y) Given that i, j are unit vectors in the positive direction of x and y axes, let vector a=(x-root number 3) i+yj, vector b=(x+root number 3) i+yj, and satisfy direction |a||b|=4 Finding the Trajectory Equation of Point P (x, y)

Method 1: using the first definition of an ellipse
Given: a =(x -√3, y), b =(x +√3, y)
And | a |+| b |=4
Then √[(x-√3)2+ y2][(x +√3)2+ y2]=4
According to the first definition of an ellipse,
It can be seen that the sum of the distances from the moving point (x, y) to the two fixed points (√3,0),(-√3,0) is 4
Then 2a=4, c=√3
By b2= a2- c2
B =1
Therefore, the trajectory equation of point P (x, y) is x2/a2+y2/b2=1
X2/4+ y2=1
Method 2: direct reduction
√[(X-√3)2+ y2][(x +√3)2+ y2]=4
√[(X-√3)2+ y2]=4-√[(x-√3)2+ y2]
Square of both sides:
(X-√3)2+ y2=16-8√[(x +√3)2+ y2]+(x +√3)2+ y2
Simplify:
2√[(X +√3)2+ y2]=√3 x +4
Square of both sides:
4[(X +√3)2+ y2]=(√3x +4)2
Simplify:
X2+4y2=4

Given that i, j are unit vectors in the positive direction of the x and y axes, let vector a=(x-root 3) i+yj, b=(x+root 3 Given that i, j are unit vectors in the positive direction of the x and y axes, let vector a=(x-root number 3) i+yj, vector b=(x+root number 3) i+yj, and satisfy the module of vector b point multiplication vector j=vector a. Finding the Trajectory Equation of Point P (x, y) The straight line l passing through the point (root number 3,0) intersects the above trajectory at two points A and B, and the modulus of AB=8 times the root number 3, so as to obtain the equation of straight line l.

(1)|(X-√3) i+yj|=√(( x-√3)2+y2)=x+√3
Y2=4√3x, parabola
(2)(√3,0) Is the focal point of the parabola. Let y=ax-√3a A (x1, y1) B (x2, y2)
With parabola, a2 x2-(2a2√3+4√3) x+3a2=0
Then x1+x2=2a2√3+4√3
Because x1+x2+2√3=8√3
So 2a2√3+4√3=6√3, i.e. a2=1
So y=x-√3, or y=-x 3

The film of vector a =1, the film of vector b = root number 3, the sum of the two vectors =(root number 3,1).

Given: vector a, b,|a|=1,|b|=√3, a+b=(√3,1);
Find:(1)|a-b|;(2) the angle α between a+b and a-b.
(1) According to the title: a2=1, b2=3,(a+b)2=4;
Let a=(x1, y1), b=(x2, y2), then a+b=(x1+x2, y1+y2)=(√3,1);
Then x12+y12=1, x22+y22=3, x1+x2=√3, y1+y2=1;
(A+b)2=(x1+x2)2+(y1+y2)2=(x12+x22)+(y12+y22)+(2x1x2+2y1y2)
=1+3+(2X1x2+2y1y2)=4, then (2x1x2+2y1y2)=0;
And a-b=(x1-x2, y1-y2),
Then (a-b)2=(x1-x2)2+(y1-y2)2
=(X12+x22+y12+y22)-(2x1x2+2y1y2)
=1+3-0=4;
Then |a-b|=2;
(2) Cosα=(a+b)·(a-b)/(|a+b||a-b|)
=(A2-b2)/(|a+b||a-b|)
=(1-3)/(2×2)
=-1/2
∵α∈[0,π],∴α=2π/3=120°.

If the vector AB=(2,3) and the coordinate of point A is (2,3), then the coordinate of point B is Suppose the vector AB=(2,3) and the coordinate of point A is (2,3), then the coordinate of point B is

0

Vector A (-2,4) vector AB (3,-1) then vector B coordinate is

0

Point A (3,5), if vector AB=(-2,-5), then the coordinate of point B is?

0