p hyperbola. Focus at pole, directrix x=12/q. ## Answer **Note:** the denominator is cut off in the scan, so the exact curve cannot be fixed. The standard form of such papers is r = 12/(p + q cosθ) (or with sinθ); the method below gives the result for any constants p, q > 0. #### Step 1: standard form r = 12/(p + q cosθ) = (12/p) / (1 + (q/p) cosθ) Compare with r = l/(1 + e cosθ): - **e = q/p**, semi-latus rectum l = 12/p - ed = l, so d = 12/q: directrix x = 12/q (to the right of the pole) - the pole is a focus; the axis of the conic is the x-axis #### Step 2: identify the conic | Condition | e | Conic | |---|---|---| | q < p | < 1 | Ellipse | | q = p | = 1 | Parabola | | q > p | > 1 | Hyperbola | (If the sign is p − q cosθ, the curve is the mirror image in the y-axis, and the directrix is x = −12/q.) #### Step 3: vertices and size - θ = 0: r = 12/(p + q), vertex (12/(p + q), 0). - θ = π: r = 12/(p − q), vertex (−12/(p − q), 0). - θ = π/2: r = 12/p, the ends of the latus rectum (0, ±12/p). - For an ellipse (q < p): a = 12p/(p² − q²), centre (−12q/(p² − q²), 0), b² = a²(1 − e²). - For a hyperbola (q > p): a = 12p/(q² − p²), b² = a²(e² − 1). #### Step 4: sketch Mark the focus at the origin, the vertices, the latus rectum ends and the directrix; draw a closed curve for e < 1, a parabola opening to the left for e = 1 (focus at the origin, vertex (6/p, 0)), or two branches for e > 1. **Worked example (p = 3, q = 2):** r = 12/(3 + 2cosθ), e = 2/3, an ellipse with vertices (12/5, 0) and (−12, 0), a = 36/5, centre (−24/5, 0), b² = 144/5, directrix x = 6. ```text y | x = 6 .-'+'-. | ---(---F2--C--F1)-----------x -12 -9.6 -4.8 0 2.4 '-.+.-' ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is r = 12/(p + q cosθ) written in standard form? A. r = (12/q)/(1 + (p/q)cosθ), e = p/q B. r = 12/(1 + pq cosθ) C. r = (12/p)/(1 + (q/p)cosθ), e = q/p D. r = 12p/(1 + q cosθ) Answer: C. Divide by p. (Key point: Standard form) 2. What decides the conic? A. The value 12 only B. The ratio q/p compared with 1 C. The sign of p only D. The angle θ Answer: B. q
p hyperbola. (Key point: Classification) 3. Where is the directrix for the + sign? A. x = 12/q B. x = −12/q C. x = 12/p D. y = 12/q Answer: A. d = l/e = 12/q to the right of the pole. (Key point: Directrix) 4. Which gives the vertices? A. θ = π/2 and 3π/2 B. θ = 0 gives 12/p C. θ = π gives 12/(p+q) D. θ = 0 gives 12/(p+q) and θ = π gives 12/(p−q) Answer: D. Vertices lie on the axis, at θ = 0 and π. (Key point: Vertices) ## Close variants set in other exams - [Sketch polar conic r = 12/(...) (2068 Chaitra)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-12-denominator-cut-2068) — Mathematics I, chapter 3, asked 2× ## Related in Mathematics I - [Sketch polar conic r = 10cosecθ/(...)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-10cosec-denominator-cut) — Mathematics I, chapter 3 - [Sketch conic r = 12/(3+2cosθ)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-12-over-3-plus-2cos) — Mathematics I, chapter 3, asked 2× - [Sketch conic r = 12/(2−6cosθ)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-12-over-2-minus-6cos) — Mathematics I, chapter 3 - [Sketch conic r = 12/(6+2sinθ)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apolar-conic-12-over-6-plus-2sin) — Mathematics I, chapter 3 ## Related in other subjects - [3-bus load flow with Q limits (cut off)](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aload-flow-reference-load-generator-bus) — Power System Analysis II, chapter 2 - [Lateral load elements; stepwise [cut off]](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Alateral-elements-stepwise-74c) — Concrete Technology and Masonry Structure, chapter 9 - [Choose best words (grammar) 2071 Shrawan](https://www.bctengineer.com/question.md?key=semester_5%2Fcommunication_english%3Achoose-best-words-grammar-set-2071s) — Communication English, chapter 3 - [Cut-off and single-mode range, 4×2 cm](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Awaveguide-4x2cm-cutoff-single-mode-range) — Electromagnetism, chapter 6