FAD1014: Rapid-Fire Drill Pack — Parametric Equations & Conic Conversion

Objective: Convert parametric equations to Cartesian form on sight and identify conic sections instantly.
Target: 45 seconds per problem. If you stall >90 seconds, skip and mark it.
Total problems: 30
Estimated time: 30 minutes

Cheat Sheet (Memorize First)

Conic Parametric Form Key Identity Cartesian Form
Circle $x-h = r\cos t$, $y-k = r\sin t$ $\cos^2 t + \sin^2 t = 1$ $(x-h)^2 + (y-k)^2 = r^2$
Ellipse $x-h = a\cos t$, $y-k = b\sin t$ $\cos^2 t + \sin^2 t = 1$ $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$
Parabola (H) $x-h = at^2$, $y-k = 2at$ --- $(y-k)^2 = 4a(x-h)$
Parabola (V) $x-h = 2at$, $y-k = at^2$ --- $(x-h)^2 = 4a(y-k)$
Hyperbola (V) $x-h = a\sec t$, $y-k = b\tan t$ $\sec^2 t - \tan^2 t = 1$ $\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1$
Hyperbola (H) $x-h = a\tan t$, $y-k = b\sec t$ $\sec^2 t - \tan^2 t = 1$ $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$

Quick Identification:

  • $\cos/\sin$ pair with same denominator $\rightarrow$ Circle (if $a=b$) or Ellipse
  • One of $x$ or $y$ has $t^2$, other has $t$ $\rightarrow$ Parabola
  • $\sec/\tan$ pair $\rightarrow$ Hyperbola

Part A: Circle Recognition (6 problems)

Target: 30 seconds per problem.

Set A1 — Circle Identification (6 problems)

Convert to Cartesian form and identify center and radius.

  1. $x = 3\cos t$, $y = 3\sin t$
  2. $x - 2 = 5\cos t$, $y + 1 = 5\sin t$
  3. $x = \cos t - 4$, $y = \sin t + 2$
  4. $x + 3 = 2\cos t$, $y - 5 = 2\sin t$
  5. $x = 7\cos t$, $y = 7\sin t$
  6. $x = \frac{1}{2}\cos t + 1$, $y = \frac{1}{2}\sin t - 3$

Score: ___/6

Part B: Ellipse Recognition (6 problems)

Target: 45 seconds per problem.

Set B1 — Ellipse Identification (6 problems)

Convert to Cartesian form and identify center and semi-axes.

  1. $x = 4\cos t$, $y = 3\sin t$
  2. $x - 1 = 6\cos t$, $y + 2 = 2\sin t$
  3. $x = 5\cos t + 3$, $y = 2\sin t - 1$
  4. $x + 4 = \cos t$, $y - 3 = 4\sin t$
  5. $x = 2\cos t$, $y = 8\sin t$
  6. $x = 3\cos t - 2$, $y = \sin t + 5$

Score: ___/6

Part C: Parabola Recognition (6 problems)

Target: 45 seconds per problem.

Set C1 — Parabola Conversion (6 problems)

Convert to Cartesian form and identify vertex and orientation.

  1. $x = 2t^2$, $y = 4t$
  2. $x = t + 1$, $y = \frac{1}{2}t^2 - 3$
  3. $x - 2 = 3t^2$, $y + 1 = 6t$
  4. $x = 4t$, $y = 2t^2 + 5$
  5. $x + 3 = t^2$, $y - 2 = 2t$
  6. $x = 6t - 1$, $y = \frac{3}{2}t^2 + 4$

Score: ___/6

Part D: Hyperbola Recognition (6 problems)

Target: 60 seconds per problem.

Set D1 — Hyperbola Conversion (6 problems)

Convert to Cartesian form and identify center and orientation.

  1. $x = 3\sec t$, $y = 2\tan t$
  2. $x - 1 = \tan t$, $y + 2 = 4\sec t$
  3. $x = 5\sec t + 2$, $y = 3\tan t - 1$
  4. $x = 2\tan t$, $y = 6\sec t$
  5. $x + 3 = \sec t$, $y - 4 = 2\tan t$
  6. $x = 4\tan t - 1$, $y = \sec t + 5$

Score: ___/6

Part E: Mixed Conics — Rapid Identification (6 problems)

Target: 60 seconds per problem.

Set E1 — Identify and Convert (6 problems)

Identify the conic type, convert to Cartesian form, and state key features.

  1. $x = \sin t + 2$, $y = \cos t - 3$
  2. $x = t^2 + 1$, $y = 2t - 4$
  3. $x = 4\cos t$, $y = 4\sin t$
  4. $x = \sec t - 2$, $y = 3\tan t + 1$
  5. $x = 6t$, $y = 3t^2$
  6. $x = 2\cos t + 5$, $y = 5\sin t - 2$

Score: ___/6

Final Scorecard

Part Sets Problems Raw Score
A — Circle A1 6 ___/6
B — Ellipse B1 6 ___/6
C — Parabola C1 6 ___/6
D — Hyperbola D1 6 ___/6
E — Mixed E1 6 ___/6
TOTAL 30 ___/30

Proficiency Benchmarks

  • 21/30 (70%) — Proficient. You can handle standard exam problems.
  • 25/30 (83%) — Solid. Fast and accurate.
  • 28/30 (93%) — Exam-ready. Any mistake is a careless slip.

Speed Benchmarks

  • <20 minutes: Excellent mechanical fluency.
  • 20-30 minutes: Good. Review missed patterns.
  • >30 minutes: Drill the specific sets you scored lowest on again tomorrow.

Error Log Template

After grading, list every wrong problem number with a one-word reason:

Problem Reason
e.g. 4 sign error

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

  1. $x^2 + y^2 = 9$; Center $(0,0)$, Radius $3$
  2. $(x-2)^2 + (y+1)^2 = 25$; Center $(2,-1)$, Radius $5$
  3. $(x+4)^2 + (y-2)^2 = 1$; Center $(-4,2)$, Radius $1$
  4. $(x+3)^2 + (y-5)^2 = 4$; Center $(-3,5)$, Radius $2$
  5. $x^2 + y^2 = 49$; Center $(0,0)$, Radius $7$
  6. $(x-1)^2 + (y+3)^2 = \frac{1}{4}$; Center $(1,-3)$, Radius $\frac{1}{2}$
  7. $\frac{x^2}{16} + \frac{y^2}{9} = 1$; Center $(0,0)$, $a=4$, $b=3$
  8. $\frac{(x-1)^2}{36} + \frac{(y+2)^2}{4} = 1$; Center $(1,-2)$, $a=6$, $b=2$
  9. $\frac{(x-3)^2}{25} + \frac{(y+1)^2}{4} = 1$; Center $(3,-1)$, $a=5$, $b=2$
  10. $\frac{(x+4)^2}{1} + \frac{(y-3)^2}{16} = 1$; Center $(-4,3)$, $a=1$, $b=4$
  11. $\frac{x^2}{4} + \frac{y^2}{64} = 1$; Center $(0,0)$, $a=2$, $b=8$
  12. $\frac{(x+2)^2}{9} + \frac{(y-5)^2}{1} = 1$; Center $(-2,5)$, $a=3$, $b=1$
  13. $y^2 = 8x$; Vertex $(0,0)$, opens right, $a=2$
  14. $(y+3)^2 = 2(x-1)$; Vertex $(1,-3)$, opens right, $a=\frac{1}{2}$
  15. $(y+1)^2 = 12(x-2)$; Vertex $(2,-1)$, opens right, $a=3$
  16. $(x)^2 = 8(y-5)$; Vertex $(0,5)$, opens up, $a=2$
  17. $(y-2)^2 = 4(x+3)$; Vertex $(-3,2)$, opens right, $a=1$
  18. $(x+1)^2 = 12(y-4)$; Vertex $(-1,4)$, opens up, $a=3$
  19. $\frac{y^2}{4} - \frac{x^2}{9} = 1$; Center $(0,0)$, opens vertically, $a=3$, $b=2$
  20. $\frac{(y+2)^2}{16} - \frac{(x-1)^2}{1} = 1$; Center $(1,-2)$, opens vertically, $a=1$, $b=4$
  21. $\frac{(y+1)^2}{9} - \frac{(x-2)^2}{25} = 1$; Center $(2,-1)$, opens vertically, $a=5$, $b=3$
  22. $\frac{y^2}{36} - \frac{x^2}{4} = 1$; Center $(0,0)$, opens vertically, $a=2$, $b=6$
  23. $\frac{(y-4)^2}{4} - \frac{(x+3)^2}{1} = 1$; Center $(-3,4)$, opens vertically, $a=1$, $b=2$
  24. $\frac{(y-5)^2}{1} - \frac{(x+1)^2}{16} = 1$; Center $(-1,5)$, opens vertically, $a=4$, $b=1$
  25. $(x-2)^2 + (y+3)^2 = 1$; Circle, Center $(2,-3)$, Radius $1$
  26. $(y+4)^2 = 4(x-1)$; Parabola, Vertex $(1,-4)$, opens right
  27. $x^2 + y^2 = 16$; Circle, Center $(0,0)$, Radius $4$
  28. $\frac{(y-1)^2}{9} - \frac{(x+2)^2}{1} = 1$; Hyperbola, Center $(-2,1)$, opens vertically
  29. $x^2 = 12y$; Parabola, Vertex $(0,0)$, opens up
  30. $\frac{(x-5)^2}{4} + \frac{(y+2)^2}{25} = 1$; Ellipse, Center $(5,-2)$, $a=2$, $b=5$

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