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Conics in polar form

Problem 8.320 · medium

For the conic \( \displaystyle r = \frac{1}{2 - 3 \cos{\left(\theta \right)}} \), 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).
\[ \frac{1}{2 \left(1 - \frac{3 \cos{\left(\theta \right)}}{2}\right)} = \frac{1}{2 - 3 \cos{\left(\theta \right)}} \]
The standard form.✓ Proved
\[ \frac{\left|{-3}\right|}{2} = \frac{3}{2} \]
e = 3/2: > 1, a hyperbola.✓ Proved
\[ 1 \cdot \frac{1}{2} \frac{1}{3 \cdot \frac{1}{2}} = \frac{1}{3} \]
ed = 1/2, so d = 1/3.✓ Proved

Answer \( e = \frac{3}{2},\ \text{hyperbola},\ \text{directrix } x = -1/3 \)

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: pass
qwen3.6:27b-mlx: fail (error) — The solution fails to determine the correct sign of the directrix. For the form r = ed / (1 - e cos theta), the directrix is x = -d. Since d = 1/3, the directrix is x = -1/3, but the derivation in step 4 only calculates the magnitude d without linking the negative sign in the denominator to the position of the directrix.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution fails to determine the correct sign of the directrix. For the form r = ed / (1 - e cos theta), the directrix is x = -d. Since d = 1/3, the directrix is x = -1/3, but the derivation in step 4 only calculates the magnitude d without linking the negative sign in the denominator to the position of the directrix.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The directrix is incorrectly identified as x = -1/3. Since the standard form has a positive cosine term (1 + e cos(theta) after rearranging signs or recognizing the form r = ed/(1 - e cos theta) implies the directrix is x = -d), the directrix should be x = -1/3 only if the form is 1 - e cos(theta). Here, r = (3/2)(1/3) / (1 - (3/2)cos(theta)), which corresponds to a directrix at x = -d = -1/3. Wait, let's re-verify. Standard form: r = ed / (1 - e cos theta) -> directrix x = -d. r = ed / (1 + e cos theta) -> directrix x = d. The equation is r = 1/(2 - 3cos theta) = (1/2) / (1 - (3/2)cos theta). So e = 3/2, ed = 1/2 => d = 1/3. The form is 1 - e cos theta, so the directrix is x = -d = -1/3. The stated answer says x = -1/3. The solution calculates d=1/3 but does not explicitly state the sign/direction logic for the directrix, just 'd=1/3'. However, the stated answer in the problem description is 'directrix x = -1/3'. The solution line 4 just says 'd=1/3'. It fails to conclude the directrix equation x = -1/3. It leaves the student with just 'd'. This is misleading/incomplete.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The directrix is identified as x = -1/3, but the standard form derived (with +1 in the denominator) corresponds to a directrix at x = +d (x = 1/3).
gpt-oss:20b: fail (error) 2026-10-14 — The solution incorrectly identifies the directrix. For the polar equation r = ed/(1+e cosθ) with e>1, the directrix is x = d, not x = –d. Thus the stated directrix x = –1/3 is wrong.

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 2026-10-17 with SymPy 1.14.0.