Asked by Jen
What is the rectangular equivalence to the parametric equations?
x(θ)=4cosθ+2,y(θ)=2sinθ−5 , where 0≤θ<2π .
___+___ = 1, where x is on the interval ___
Options: (x-2)^2 /16, (x-2)^2 /16, (x-5)^2 / 4, (x+5)^2 /4, (y-2)^5/16,
(y+2)^5 /16, (y-5)^2 /4, (y+5)^2 /4, [-7,-3],[-4,4]. [-2,6], [-2,2]
x(θ)=4cosθ+2,y(θ)=2sinθ−5 , where 0≤θ<2π .
___+___ = 1, where x is on the interval ___
Options: (x-2)^2 /16, (x-2)^2 /16, (x-5)^2 / 4, (x+5)^2 /4, (y-2)^5/16,
(y+2)^5 /16, (y-5)^2 /4, (y+5)^2 /4, [-7,-3],[-4,4]. [-2,6], [-2,2]
Answers
Answered by
oobleck
(x-2)/4 = cosθ
(y+5)/2 = sinθ
now, since cos^2θ + sin^2θ = 1, ...
(y+5)/2 = sinθ
now, since cos^2θ + sin^2θ = 1, ...
Answered by
Reiny
x = 4cosθ + 2 ---> cosθ = (x-2)/4
y = 2sinθ - 5 -----> sinθ = (y+5)/2
we know sin^2 θ + cos^2 θ = 1
(y+5)^2 /4 + (x-2)^2/16 = 1
(x-2)^2/16 + (y+5)^2 /4 = 1
an ellipse with centre at (2, -5) , |a| = 4, |b| = 2
check:
https://www.wolframalpha.com/input/?i=plot+x%3D4cos%CE%B8%2B2%2C+y%3D2sin%CE%B8%E2%88%925
y = 2sinθ - 5 -----> sinθ = (y+5)/2
we know sin^2 θ + cos^2 θ = 1
(y+5)^2 /4 + (x-2)^2/16 = 1
(x-2)^2/16 + (y+5)^2 /4 = 1
an ellipse with centre at (2, -5) , |a| = 4, |b| = 2
check:
https://www.wolframalpha.com/input/?i=plot+x%3D4cos%CE%B8%2B2%2C+y%3D2sin%CE%B8%E2%88%925
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