Given the vector PA (3, cos α), Pb = (sin α, 1) and PA ⊥ Pb, (1) find the value of Tan α, sin α, cos α, (2) find ì ab ì

Given the vector PA (3, cos α), Pb = (sin α, 1) and PA ⊥ Pb, (1) find the value of Tan α, sin α, cos α, (2) find ì ab ì

(1) Because PA ⊥ PB
So PA * Pb = 0
That is, 3sin α + cos α = 0
3sinα=-cosα
Easy to get Tan α = - 1 / 3
So α is the second or fourth quadrant angle
And Sin & # 178; α + cos & # 178; α = 1
So Sin & # 178; α + 9sin & # 178; α = 1
The results show that sin & # 178; α = 1 / 10
When α is the second quadrant angle, sin α = √ 10 / 10, cos α = - 3 √ 10 / 10
When α is the fourth quadrant angle, sin α = - 10 / 10, cos α = 3 √ 10 / 10
(2) Because the vector AB = pb-pa and PA * Pb = 0
So | ab | & # 178; = | pb-pa | & # 178;
=|PB|²-2PB*PA+|PA|²
=9+cos²α+sin²α+1
=11
So | ab | = √ 11

It is known that the eccentricity of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is √ 3 / 2,
If AF = 3fb, then k =?

E = √ 3 / 2 → C / a = √ 3 / 2 → B / a = 1 / 2 ellipse C: x ^ 2 / 4B ^ 2 + y ^ 2 / b ^ 2 = 1 let line AB: my = x - C = x - √ 3B let a (x1, Y1), B (X2, Y2) AF / BF = 3 → Ly1 / y2l = 3 let line AB substitute ellipse x ^ 2 + 4Y ^ 2 = 4B ^ 24B ^ 2 = (my + √ 3b) ^ 2 + 4Y ^ 2 = (m ^ 2 + 4)

It is known that the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) passes through the point (1,3 / 2), and the eccentricity e = 1 / 2 (1) the standard equation of ellipse C is x ^ 2 / 4 + y ^ 2 / 3 = 1 (2). If the line L: y = KX + m (K ≠ 0) intersects the ellipse with different two points m, N, and the vertical bisector of line Mn passes through the point G (1 / 8,0), the value range of K is obtained

Substitute the point (1,3 / 2) into the ellipse to get: 1 / A ^ 2 + 9 / 4B ^ 2 = 1
E = C / a = 1 / 2, then a = 2c, and a ^ 2 = B ^ 2 + C ^ 2, the solution is a ^ 2 = 4, B ^ 2 = 3
So the elliptic C equation is: x ^ 2 / 4 + y ^ 2 / 3 = 1
Let m (x1, Y1), n (X2, Y2) and the midpoint of Mn be p, then p ((x1 + x2) / 2, (Y1 + Y2) / 2)
Substituting the straight line L: y = KX + m into the elliptic equation, we get: (4K ^ 2 + 3) x ^ 2 + 8kmx + 4m ^ 2-12 = 0
x1+x2=-8km/（4k^2+3),y1+y2=k(x1+x2)+2m=6m/（4k^2+3)
Δ > 0: 4K ^ 2-m ^ 2 + 3 > 0 is m ^ 2

Let t be a parameter and y = TX, try to change y = 4x ^ 2-5x ^ 3 into a parametric equation

x=(2+(4-5t)^(1/2))/5
y=(2t+(4-5t)^(1/2)*t)/5
or
x=(2-(4-5t)^(1/2))/5
y=(2t-(4-5t)^(1/2)*t)/5