Answer:
[tex](y+3)^2-\dfrac{(x-2)^2}{16}=1[/tex]
Step-by-step explanation:
You want the system of parametric equations (x, y) = (2-4tan(2t), sec(2t) -3) rewritten as one equation in x and y.
Trig identity
We can solve for tan(2t) and sec(2t) and use the trig identity ...
sec(2t)² = tan(2t)² +1
Trig expressions
y +3 = sec(2t) . . . . . . add 3
x -2 = -4tan(2t) . . . . . subtract 2
(x -2)/(-4) = tan(2t) . . . . divide by 4
Using these expressions in the above identity, we get ...
[tex](y+3)^2 = \left(\dfrac{x-2}{-4}\right)^2 +1[/tex]
Subtracting the x-term gives the equation of a hyperbola that opens along the y-axis:
[tex]\boxed{(y+3)^2-\dfrac{(x-2)^2}{16}=1}[/tex]
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