If the equation x ^ 2sina-y ^ 2cosa = 1 (0 ≤ a

If the equation x ^ 2sina-y ^ 2cosa = 1 (0 ≤ a

A:
x^2sina-y^2cosa=1(0≤a-1/cosa>0
So:
-cosa>sina>0,cosasina
So: 3 π / 4

The equation x2sin α + y2cos α = 1 represents a hyperbola with the focus on the y-axis, then the angle α is on the second axis______ Quadrant

∵ equation x2sin α + y2cos α = 1 means hyperbola with focus on y-axis, ∵ cos α > 0, sin α < 0, angle α takes points (x, y) on the terminal edge of angle α, then x > 0, y < 0, angle α is in the fourth quadrant, so the answer is: four

If the equation x ^ 2sina-y ^ 2cosa = 1 (0 ≤ a ＜ 2 π) denotes an ellipse, it is an exercise to find the value range of A

It can be seen from the question
Sina > 0 and cosa < 0
∴π/2＜a＜π

Through the point P (- 3,3), make a straight line L intersection ellipse x + 2cos α, y + sin α (α is the parameter) at two points a and B. If | PA | * | Pb | = 164 / 7, find the equation of the straight line

Ellipse x = 2cos α, y = sin α
==>x/2=cosα,y=sinα
Sum of squares:
x^2/4+y^2=1 (1)
Let the parameter equation of the straight line be:
x=-3+tcosθ,y=3+tsinθ
(t is the parameter and θ is the tilt angle constant)
Substituting (1)
（-3+tcosθ)^2+4(3+tsinθ)^2=4
(cos²θ+4sin²θ)t²+(24sinθ-6cosθ)t+41=0
Δ=(24sinθ-6cosθ)²-164 (cos²θ+4sin²θ)
=-180sin²θ-28cos²θ-288sinθcosθ>0
According to Weida's theorem:
t1*t2=41/(cos²θ+4sin²θ)
∵|PA|*|PB|=164/7=|t1t2|
∴41/(cos²θ+4sin²θ)=164/7
∴cos²θ+4sin²θ=7/4
∴1-sin²θ+4sin²θ=7/4
Ψ Sin & # 178; θ = 1 / 4 = = > sin θ = 1 / 2 (rounding off)
cos²θ=3/4 ==> cosθ=±√3/2
∵Δ>0 ∴sinθcosθ