What is the area of the parallelogram shown?please help!!!!
1 answer:
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Answer:
α(t) = (-a*Sin(t), b*Cos (t)) where t ∈ [0, 2π]
Step-by-step explanation:
Given an arbitrary ellipse (x²/a²) + (y²/b²) = 1 (a, b > 0)
The parametrization can be as follows
x = -a*Sin(t)
y = b*Cos (t)
then
α(t) = (-a*Sin(t), b*Cos (t)) where t ∈ [0, 2π]
If t = 0
α(0) = (-a*Sin(0), b*Cos (0)) = (0, b)
If t = π/2
α(π/2) = (-a*Sin(π/2), b*Cos (π/2)) = (-a, 0)
If t = π
α(π) = (-a*Sin(π), b*Cos (π)) = (0, -b)
If t = 3π/2
α(3π/2) = (-a*Sin(3π/2), b*Cos (3π/2)) = (a, 0)
If t = 2π
α(2π) = (-a*Sin(2π), b*Cos (2π)) = (0, b)
We can see the sketch in the pic.
