Hyperbola x2/a2-y2/b2=1(a >0, b >0) is known to be directed by vector k =(6,6) If the midpoint of the chord cut by the straight line of is (4,1), then the value of the hyperbolic eccentricity is A.√5/2 B.√6/2 C.√10/3 D.2

Hyperbola x2/a2-y2/b2=1(a >0, b >0) is known to be directed by vector k =(6,6) If the midpoint of the chord cut by the straight line of is (4,1), then the value of the hyperbolic eccentricity is A.√5/2 B.√6/2 C.√10/3 D.2

Let A (x1, y1), B (x2, y2) be the intersection points of straight line and hyperbola respectively, then x1^2/a^2-y1^2/b^2=1, x2^2/a^2-y2^2/b^2=1, subtract (x2+x1)(x2-x1)/a^2-(y2+y1)(y2-y1)/b^2=0. Since the midpoint of AB is (4,1), x1+x2=8, y1+y2=2, substitute 8...

Given △ three internal angles A, B, C, vector m=(4,1), n=(sin2A/2, cos2A), and m·n=1/2 Find Angle A Size 4Sin2A/2+ cos2A=1/2 . 4 Cos2A -8 cosA +5=0 What's wrong with me?

4Sin2A/2+ cos2A=1/2
4×(1-CosA)/2+2 cos2A -1=1/2
4-4CosA+4cos2A-2=1
4 Cos2A -4 cosA +1=0
(2 CosA-1)2=0
CosA=1/2
A is three internal angles
A =60°

4Sin2A/2+ cos2A=1/2
4×(1-CosA)/2+2 cos2A -1=1/2
4-4CosA+4cos2A-2=1
4 Cos2A -4 cosA +1=0
(2 CosA-1)2=0
CosA=1/2
A is three inside corners
A =60°

Given vector |a |2=|b |2=1a multiplied by b =-1/2 1. Find |a+b|2. Find the included angle between a and b-a

|A|=|b|=1
A*b=-1/2
(1)
|A+b|2=(a+b)2=a2+2a*b+b2=|a|2+2a*b+|b|2=1+2*(-1/2)+1=1
(2)
|B-a|2=(b-a)2=b2-2a*b+a2=|b|2-2a*b+|a|2=1-2*(-1/2)+1=3
So |b-a|=√3
A*(b-a)=a*b-a2=a*b-|a|2=-1/2-1=-3/2
Let the angle between a and b-a be θ
Then a*(b-a)=|a||b-a|*cosθ=1 3*cosθ=-3/2
So cos θ=-√3/2
So θ=150°

Given vector a (2,1,3) vector b (-4,2,2) and vector a perpendicular vector b, then |vector a-vector b |=___

Vector a Vertical vector b
Then | vector a-vector b |^2=a^2+b^2=14+24=38
Therefore|vector a-vector b|=√34

Vector a Vertical vector b
Then |vector a-vector b |^2=a^2+b^2=14+24=38
Therefore|vector a-vector b|=√34

Given vector a=(n,1) and vector b=(4, n), vector a.b is collinear when finding the value of n and (2a+b) is perpendicular to b when finding the value of n

A, b collinear, there must be X1Y2-X2Y1=0, that is, n square-1X4=0, n=positive and negative 2, when n=-2, a, b direction is opposite
Let k=(2a+b) then k=2x (n,1)+(4, n)=(2n+4, n+2)
K is perpendicular to b, with X1X2+Y1Y2=0, i.e.(2n+4) x4+(n+2) xn=0
Solution, n=-2 or -8(k is 0 vector when n=-2)

Verification: A triangle consists of three midlines of a triangle is connected end to end to form a triangle Verification: the vector consisting of three midlines of a triangle is connected end to end to form a triangle

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