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Sagot :
Answer:
C, A, E, B, D
Step-by-step explanation:
The conversion relations between rectangular and polar coordinates can be used to find the polar coordinate equivalents. Those relations are ...
- x = r·cos(θ)
- y = r·sin(θ)
- x² +y² = r²
__
x = 2
Using the expression for x, this is ...
r·cos(θ) = 2
r = 2/cos(θ) . . . . divide by cos(θ)
r = 2·sec(θ) . . . . use the trig identity
__
x²+y²=36
From above, this becomes ...
r² = 36
r = 6 . . . . . . take the square root
__
x²+y²=2y
From above, this becomes ...
r² = 2·r·sin(θ)
r = 2·sin(θ) . . . . . . divide by r
__
x=√3y
Using the above relations, this is ...
r·cos(θ) = √3·r·sin(θ)
1/√3 = sin(θ)/cos(θ) . . . . . . divide by √3·r·cos(θ)
tan(θ) = 1/√3 ⇒ θ = arctan(1/√3)
θ = π/6
__
x=y
From above, this is ...
r·cos(θ) = r·sin(θ)
1 = sin(θ)/cos(θ) = tan(θ) . . . . divide by r·cos(θ)
θ = arctan(1)
θ = π/4
_____
Additional comment
An equation of the form y = kx defines a line through the origin and through opposite quadrants of the Cartesian plane. Above, we found the equivalent polar equation to be θ = arctan(k). Using the principal branch of the arctangent function, this is the equation of a ray.
However, the tangent function is periodic with period π, so θ = arctan(k)+π is also an equivalent to the rectangular equation. It is the opposite ray, so forms the complete line when joined with the first ray.
The correct answer for the polar coordinates will be C, A, E, B, D
What is a polar coordinate system?
The polar coordinate system is a two-dimensional coordinate system in which a distance from a reference point and an angle from a reference direction identify each point on a plane. The pole is the reference point, and the polar axis is the ray from the pole in the reference direction.
The conversion relations between rectangular and polar coordinates can be used to find the polar coordinate equivalents. Those relations are ...
x = r·cos(θ)
y = r·sin(θ)
x² +y² = r²
x = 2
Using the expression for x, this is ...
r·cos(θ) = 2
r = 2/cos(θ) . . . . divide by cos(θ)
r = 2·sec(θ) . . . . use the trig identity
x²+y²=36
From above, this becomes ...
r² = 36
r = 6 . . . . . . take the square root
x²+y²=2y
From above, this becomes ...
r² = 2·r·sin(θ)
r = 2·sin(θ) . . . . . . divide by r
x=√3y
Using the above relations, this is ...
r·cos(θ) = √3·r·sin(θ)
1/√3 = sin(θ)/cos(θ) . . . . . . divide by √3·r·cos(θ)
tan(θ) = 1/√3 ⇒ θ = arctan(1/√3)
θ = π/6
x=y
From above, this is ...
r·cos(θ) = r·sin(θ)
1 = sin(θ)/cos(θ) = tan(θ) . . . . divide by r·cos(θ)
θ = arctan(1)
θ = π/4
However, the tangent function is periodic with period π, so θ = arctan(k)+π is also equivalent to the rectangular equation. It is the opposite ray, so forms the complete line when joined with the first ray.
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