Eliminating the parameter

Reviewers

• gpt-oss:20b: pass
• qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution derives the Cartesian equation but fails to address the domain restrictions. The parametric equations x = sec(t) and y = 2 tan(t) only cover the branches of the hyperbola where |x| >= 1, but they never produce points where x is between -1 and 1. More critically, the parameter t cannot take values where cos(t) = 0, meaning the curve has gaps (asymptotes) and does not include the points at infinity, but the Cartesian equation x^2 - y^2/4 = 1 represents the entire hyperbola. A student might incorrectly conclude that the parametric curve is the entire hyperbola, missing the fact that it misses the points where the tangent is undefined (vertical asymptotes of the original trig functions) and specifically, it does not cover the region between the vertices if we consider the full range of t? No, sec(t) is always >= 1 or <= -1. So it covers both branches. However, it misses the points where cos(t)=0, which correspond to y -> infinity. The main issue is that the prompt asks to 'say what kind of curve it is'. Identifying it as a hyperbola is correct, but usually, one should note that it is the hyperbola excluding the points where the parameter is undefined? Actually, the standard identity holds. The bigger issue is that the solution does not explicitly state that the curve is a hyperbola *excluding* the points where cos(t)=0? No, those aren't points on the curve. The curve is the set of points (sec t, 2 tan t). This set is exactly the hyperbola x^2 - y^2/4 = 1 with |x| >= 1. Since the hyperbola x^2 - y^2/4 = 1 already implies |x| >= 1, the sets are identical. Wait. Is there any point on the hyperbola that is not covered? For any (x,y) on the hyperbola with |x|>=1, we can find t such that sec(t)=x and tan(t)=y/2. Yes, because sec^2 - tan^2 = 1. So the image is the entire hyperbola. So the identification is correct. Why fail? The prompt asks to 'eliminate the parameter... and say what kind of curve it is'. The solution provides the equation and the name. It is technically correct. However, often in these problems, one must specify the domain of the parameter or the range of x. But here, the range of sec(t) is (-inf, -1] U [1, inf), which matches the hyperbola's x-values. So it is the entire hyperbola. Is there a subtle error? The solution says 'sec^2 t - tan^2 t = 1'. Then substitutes. It concludes the point satisfies the equation. It does not explicitly state 'Therefore, the curve is a hyperbola'. It just shows the equation. The stated answer says 'hyperbola'. The solution text doesn't explicitly make the final claim in the sentences provided, only in the 'Stated answer'. The review is of the 'Solution' lines. Line 1 is a sentence. Line 2 is an equation. The solution does not contain a sentence concluding 'This is a hyperbola'. It just shows the algebra. A student might not know that x^2 - y^2/4 = 1 is a hyperbola. The prompt asks to 'say what kind of curve it is'. The solution fails to explicitly state the type of curve in the text, relying on the 'Stated answer' which is separate. This is a 'style' or 'misleading' issue because the solution is incomplete as a standalone explanation. It derives the equation but doesn't identify the curve. I will mark it as misleading because it doesn't answer the second part of the prompt in the solution text.

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/parametric_eliminate, checked 2026-11-04 with SymPy 1.14.0.