Conics in polar form
Problem 8.432 · medium
For the conic \( \displaystyle r = \frac{10}{1 - 2 \cos{\left(\theta \right)}} \), find the eccentricity, identify the conic, and find the directrix.
- Divide top and bottom by the constant in the denominator to get r = ed/(1 ± e cos θ).
- \[ \frac{10}{1 \left(- \frac{2 \cos{\left(\theta \right)}}{1} + 1\right)} = \frac{10}{1 - 2 \cos{\left(\theta \right)}} \]The standard form.✓ Proved
- \[ \frac{\left|{-2}\right|}{1} = 2 \]e = 2: > 1, a hyperbola.✓ Proved
- \[ \frac{10}{1 \cdot 2} = 5 \]ed = 10, so d = 5.✓ Proved
Answer \( e = 2,\ \text{hyperbola},\ \text{directrix } x = -5 \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. 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 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ 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 | for points on the curve, distance to the focus over distance to the directrix is e |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution correctly identifies e=2 and d=5, but fails to determine the correct sign for the directrix. The form r = ed/(1 - e cos theta) corresponds to a directrix at x = -d, so the directrix is x = -5. While the stated answer is correct, the solution text does not derive the sign, leaving a gap in the logic regarding the position of the directrix relative to the pole.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (style) 2026-10-18 — [domain objection, downgraded to style] The solution correctly identifies e=2 and d=5, but fails to determine the correct sign for the directrix. The form r = ed/(1 - e cos theta) corresponds to a directrix at x = -d, so the directrix is x = -5. While the stated answer is correct, the solution text does not derive the sign, leaving a gap in the logic regarding the position of the directrix relative to the pole.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution correctly identifies e=2 and d=5, but fails to determine the correct sign for the directrix. The form r = ed/(1 - e cos theta) corresponds to a directrix at x = -d, so the directrix is x = -5. While the stated answer is correct, the solution steps do not justify the negative sign, leaving a gap in the logic that a student might miss.gpt-oss:20b: fail (misleading) 2026-10-16 — The solution incorrectly concludes the directrix is x = -5. For the polar equation r = ed/(1 - e cosθ) with e = 2 and d = 5, the directrix is the line x = +5, not x = -5. The sentence claiming x = -5 misleads a student about the relationship between the sign in the denominator and the location of the directrix.qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution fails to account for the sign in the denominator. The form r = ed / (1 - e cos θ) corresponds to a directrix at x = -d, not x = d. Since e=2 and ed=10 implies d=5, the directrix is x = -5. While the final answer happens to be correct, the reasoning in step 4 (implying d=5 directly determines the directrix without addressing the sign convention) is flawed and would lead to errors in other cases (e.g., if the denominator were 1 + 2 cos θ). A rigorous solution must explicitly link the minus sign in the denominator to the negative x-coordinate of the directrix.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/polar_conics, checked 2026-10-18 with SymPy 1.14.0.