To eliminate the parameter $$t$$, we isolate the trigonometric functions. From $$x = 5 + 2\cos(t)$$, we get $$\cos(t) = \frac{x-5}{2}$$. From $$y = -1 + 6\sin(t)$$, we get $$\sin(t) = \frac{y+1}{6}$$. Using the identity $$\cos^2(t) + \sin^2(t) = 1$$, we substitute to get $$(\frac{x-5}{2})^2 + (\frac{y+1}{6})^2 = 1$$, which simplifies to $$\frac{(x-5)^2}{4} + \frac{(y+1)^2}{36} = 1$$.

The standard parametrization for a vertical hyperbola $$\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1$$ is $$x(t) = h + a\tan(t)$$ and $$y(t) = k + b\sec(t)$$. Here, the center is $$(h,k) = (1, -3)$$, $$a = \sqrt{36} = 6$$, and $$b = \sqrt{4} = 2$$. So, $$x(t) = 1 + 6\tan(t)$$ and $$y(t) = -3 + 2\sec(t)$$.

The standard parametrization for an ellipse $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$ is $$x(t) = h + a\cos(t)$$ and $$y(t) = k + b\sin(t)$$. Here, the center is $$(h,k) = (-1, 4)$$. The horizontal semi-axis is $$a = \sqrt{9} = 3$$, and the vertical semi-axis is $$b = \sqrt{25} = 5$$. So, $$x(t) = -1 + 3\cos(t)$$ and $$y(t) = 4 + 5\sin(t)$$.

From $$x(t) = e^t$$, we can see that $$x$$ must be positive. We can also write $$e^{2t} = (e^t)^2 = x^2$$. Substituting this into the equation for $$y(t)$$ gives $$y = 2x^2 - 1$$. Since $$x=e^t$$, the domain of the rectangular equation is restricted to $$x>0$$.

To identify the conic, we eliminate the parameter $$t$$. From the given equations, we have $$\tan(t) = \frac{x+2}{3}$$ and $$\sec(t) = \frac{y-1}{5}$$. Using the Pythagorean identity $$\sec^2(t) - \tan^2(t) = 1$$, we get $$\left(\frac{y-1}{5}\right)^2 - \left(\frac{x+2}{3}\right)^2 = 1$$. This equation, $$\frac{(y-1)^2}{25} - \frac{(x+2)^2}{9} = 1$$, is the standard form of a hyperbola.

The standard parametrization for a circle with center $$(h, k)$$ and radius $$r$$ is $$x(t) = h + r\cos(t)$$ and $$y(t) = k + r\sin(t)$$. The given equation has center $$(3, -2)$$ and radius $$r = \sqrt{16} = 4$$. Therefore, the correct parametrization is $$x(t) = 3 + 4\cos(t)$$, $$y(t) = -2 + 4\sin(t)$$ for a full cycle, $$0 \le t < 2\pi$$.

The standard counter-clockwise parametrization for a circle of radius 3 is $$x(t) = 3\cos(t), y(t) = 3\sin(t)$$. The top half of the circle corresponds to $$y \ge 0$$. The function $$y(t) = 3\sin(t)$$ is non-negative for $$0 \le t \le \pi$$. Therefore, this domain for $$t$$ traces the top half of the circle.

This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on hyperbola parametrization using trigonometric identities. Parametrization involves expressing implicitly defined functions through parameters, with hyperbolas requiring the hyperbolic identity sec²(t) - tan²(t) = 1. In this scenario, the equation x² - y² = 1 represents a hyperbola, which cannot use the circular identity cos²(t) + sin²(t) = 1. Choice B is correct because x = sec(t) and y = tan(t) satisfy the hyperbola equation: sec²(t) - tan²(t) = 1, which is a fundamental trigonometric identity. Choice A is incorrect because cos²(t) - sin²(t) = cos(2t), not 1, so this parametrization doesn't trace the given hyperbola. To help students: Distinguish between circle/ellipse parametrizations (using sin and cos) and hyperbola parametrizations (using sec and tan or hyperbolic functions). Emphasize the identity sec²(t) - tan²(t) = 1 as the hyperbolic analogue to cos²(t) + sin²(t) = 1.

This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on ellipse parametrization from standard form. Parametrization involves expressing an implicitly defined function in terms of parameters, with ellipses requiring scaling of the basic circular parametrization. In this scenario, the ellipse x²/16 + y²/9 = 1 has semi-major axis a = 4 (horizontal) and semi-minor axis b = 3 (vertical). Choice B is correct because x = 4cos(t) and y = 3sin(t) satisfy the ellipse equation: (4cos(t))²/16 + (3sin(t))²/9 = 16cos²(t)/16 + 9sin²(t)/9 = cos²(t) + sin²(t) = 1. Choice A is incorrect because it uses 16 and 9 directly instead of their square roots, giving x = 16cos(t) and y = 9sin(t), which would trace a much larger ellipse. To help students: For an ellipse x²/a² + y²/b² = 1, the parametrization is x = a·cos(t), y = b·sin(t). Always take the square root of the denominators to find the semi-axes lengths.

This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on matrix transformations of parametric curves. Parametrization combined with matrix multiplication allows systematic transformation of curves, where diagonal matrices perform scaling operations on each component independently. In this scenario, the diagonal matrix A = [[2, 0], [0, 5]] scales the unit circle parametrization ⟨cos(t), sin(t)⟩. Choice A is correct because matrix multiplication gives: [[2, 0], [0, 5]] × [cos(t), sin(t)]ᵀ = [2cos(t), 5sin(t)]ᵀ, which represents an ellipse with horizontal semi-axis 2 and vertical semi-axis 5. Choice D is incorrect because it replaces sin(t) with cos(t) in the y-component, giving ⟨2cos(t), 5cos(t)⟩, which would trace a line segment rather than an ellipse. To help students: Emphasize that diagonal matrices scale each component independently - the (1,1) entry scales x and the (2,2) entry scales y. Practice matrix-vector multiplication step by step to avoid confusion.