Solve the system of linear equations by solving y<2x-4
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.
Because this is a positive parabola, it opens upwards, like a cup, and the vertex dictates what the minimum value of the function is. In order to determine the vertex, I recommend completing the square. Do that by first setting the function equal to 0 and then moving the 9 to the other side by subtraction. So far: 
. Now, to complete the square, take half the linear term, square it, and add that number to both sides. Our linear term is 6. Half of 6 is 3 and 3 squared is 9. So add 9 to both sides. 
. The right side reduces to 0, and the left side simplifies to the perfect square binomial we created while completing this process. 
. Move the 0 back over and the vertex is clear now. It is (-3, 0). Therefore, 0 is the minimum point on your graph. The first choice above is the one you want.