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Eliminating the parameter

Problem 8.363 · medium

Eliminate the parameter from \( \displaystyle x = 5 \sec{\left(t \right)},\ y = \tan{\left(t \right)} \), and say what kind of curve it is.

sec²t − tan²t = 1.
\[ \left. \frac{x^{2}}{25} - y^{2} - 1 \right|_{\substack{ x=5 \sec{\left(t \right)}\\ y=\tan{\left(t \right)} }} = 0 \]
The parametric point satisfies the equation for every t.✓ Proved

Answer \( \frac{x^{2}}{25} - y^{2} = 1\ (\text{hyperbola}) \)

Lines: 1 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	points on the parametric curve at four values of t satisfy the equation

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution fails to address the domain restrictions. The parametric equations define only the branches of the hyperbola where |x| >= 5, but the Cartesian equation x^2/25 - y^2 = 1 represents the entire hyperbola. Stating that the curve is simply 'a hyperbola' without noting that it is a subset of the hyperbola (specifically, the two branches excluding the segment between the vertices if t is restricted, or just the two branches if t is real) is misleading because the parameterization does not cover the entire curve defined by the implicit equation (e.g., points with |x| < 5 are not on the curve, but the implicit equation alone doesn't explicitly exclude them without context, though technically the implicit equation also doesn't have points with |x|<5. Wait, x^2/25 - y^2 = 1 implies x^2 >= 25, so |x| >= 5. So the implicit equation actually describes exactly the same set of points as the parameterization for t in R? Let's check. sec(t) ranges over (-inf, -1] U [1, inf). So x = 5 sec(t) ranges over (-inf, -5] U [5, inf). The equation x^2/25 - y^2 = 1 implies x^2 >= 25, so |x| >= 5. So the set of points is identical. However, usually 'eliminate the parameter' problems require stating the domain or noting that it is a hyperbola. Is it misleading? The prompt asks 'what kind of curve it is'. 'Hyperbola' is the correct classification. But is there a nuance? The parameterization covers the entire hyperbola? Yes, as t varies over R (excluding pi/2 + kpi), sec(t) takes all values in (-inf, -1] U [1, inf). So the image is the entire hyperbola. So the statement is actually correct. Why did I think it was misleading? Maybe because sometimes parameterizations miss points? Here it doesn't. Let's re-read carefully. 'sec^2 t - tan^2 t = 1'. This identity holds for all t where sec and tan are defined. The substitution shows the points lie on the hyperbola. Does every point on the hyperbola correspond to a t? Yes. So the curve is indeed the hyperbola. Is there any error? The solution is very brief. It states the identity, shows the substitution works. It concludes it's a hyperbola. This seems sound. Is there a 'style' issue? It's very terse. But the instructions say 'pass' if sound. Let's look for subtle errors. 'sec^2 t - tan^2 t = 1'. This is true. The substitution proves that the parametric curve is a subset of the hyperbola. It does not prove that the parametric curve is the *entire* hyperbola. To fully eliminate the parameter and identify the curve, one should ideally show surjectivity or at least mention the range. However, in standard calculus contexts, showing the implicit equation is often sufficient to identify the 'kind' of curve. But strictly speaking, the solution doesn't prove the curve *is* the hyperbola, only that it lies on it. If the parameterization was x = 5 sec(t), y = tan(t) for t in [0, pi/4], it would be an arc. The problem doesn't specify the domain of t. Usually, t is assumed to be all real numbers where defined. In that case, it is the whole hyperbola. If the domain isn't specified, assuming it's the maximal domain is standard. So it's likely 'pass'. But wait, is there a sign error? x = 5 sec t. x^2 = 25 sec^2 t. x^2/25 = sec^2 t. y = tan t. y^2 = tan^2 t. x^2/25 - y^2 = sec^2 t - tan^2 t = 1. Correct. The identification 'hyperbola' is correct. I will mark it pass.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-10-18 — [domain objection, downgraded to style] The solution fails to address the domain restrictions. The parametric equations define only the branches of the hyperbola where |x| >= 5, but the Cartesian equation x^2/25 - y^2 = 1 represents the entire hyperbola. Stating that the curve is simply 'a hyperbola' without noting that it is a subset of the hyperbola (specifically, the two branches excluding the segment between the vertices if t is restricted, or just the two branches if t is real) is misleading because the parameterization does not cover the entire curve defined by the implicit equation (e.g., points with |x| < 5 are not on the curve, but the implicit equation alone doesn't explicitly exclude them without context, though technically the implicit equation also doesn't have points with |x|<5. Wait, x^2/25 - y^2 = 1 implies x^2 >= 25, so |x| >= 5. So the implicit equation actually describes exactly the same set of points as the parameterization for t in R? Let's check. sec(t) ranges over (-inf, -1] U [1, inf). So x = 5 sec(t) ranges over (-inf, -5] U [5, inf). The equation x^2/25 - y^2 = 1 implies x^2 >= 25, so |x| >= 5. So the set of points is identical. However, usually 'eliminate the parameter' problems require stating the domain or noting that it is a hyperbola. Is it misleading? The prompt asks 'what kind of curve it is'. 'Hyperbola' is the correct classification. But is there a nuance? The parameterization covers the entire hyperbola? Yes, as t varies over R (excluding pi/2 + kpi), sec(t) takes all values in (-inf, -1] U [1, inf). So the image is the entire hyperbola. So the statement is actually correct. Why did I think it was misleading? Maybe because sometimes parameterizations miss points? Here it doesn't. Let's re-read carefully. 'sec^2 t - tan^2 t = 1'. This identity holds for all t where sec and tan are defined. The substitution shows the points lie on the hyperbola. Does every point on the hyperbola correspond to a t? Yes. So the curve is indeed the hyperbola. Is there any error? The solution is very brief. It states the identity, shows the substitution works. It concludes it's a hyperbola. This seems sound. Is there a 'style' issue? It's very terse. But the instructions say 'pass' if sound. Let's look for subtle errors. 'sec^2 t - tan^2 t = 1'. This is true. The substitution proves that the parametric curve is a subset of the hyperbola. It does not prove that the parametric curve is the *entire* hyperbola. To fully eliminate the parameter and identify the curve, one should ideally show surjectivity or at least mention the range. However, in standard calculus contexts, showing the implicit equation is often sufficient to identify the 'kind' of curve. But strictly speaking, the solution doesn't prove the curve *is* the hyperbola, only that it lies on it. If the parameterization was x = 5 sec(t), y = tan(t) for t in [0, pi/4], it would be an arc. The problem doesn't specify the domain of t. Usually, t is assumed to be all real numbers where defined. In that case, it is the whole hyperbola. If the domain isn't specified, assuming it's the maximal domain is standard. So it's likely 'pass'. But wait, is there a sign error? x = 5 sec t. x^2 = 25 sec^2 t. x^2/25 = sec^2 t. y = tan t. y^2 = tan^2 t. x^2/25 - y^2 = sec^2 t - tan^2 t = 1. Correct. The identification 'hyperbola' is correct. I will mark it pass.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (style) 2026-10-16 — [domain objection, downgraded to style] The solution identifies the curve as a hyperbola but fails to specify the domain restrictions (|x| >= 5). The parametric equations cannot produce points with |x| < 5, so the curve is not the entire hyperbola, only the two branches. Stating it is simply 'a hyperbola' without qualification is misleading in this context.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the trigonometric identity needed to eliminate the parameter and verifies that the parametric equations satisfy the resulting Cartesian equation. The identification of the curve as a hyperbola is correct.
gpt-oss:20b: pass 2026-10-15

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 2026-10-18 with SymPy 1.14.0.