- Practice · 6 marks
Parametric parabola and hyperbola points
Short answer
Parabola: x=3t^2, y=6t; points (0,0), (0.75,3), (3,6), (12,12); slope at t=1 is 1. Hyperbola: x=3secθ, y=4tanθ; points (3,0), (3.464,2.309), (4.243,4).
Write the standard parametric equations of a parabola and a hyperbola. For the parabola y^2 = 12x, calculate the points for t = 0, 0.5, 1 and 2 and the slope of the tangent at t = 1. For the hyperbola x^2/9 - y^2/16 = 1, calculate points for theta = 0, 30 and 45 degrees.
Conic sections (circle, ellipse, parabola, hyperbola) are the analytic curves used in CAD. Their parametric forms with centre or vertex at the origin are:
- Parabola : , , .
- Hyperbola : , (or , ).
Parabola
, so , .
| 0 | 0 | 0 |
| 0.5 | 0.75 | 3 |
| 1 | 3 | 6 |
| 2 | 12 | 12 |
Check at : .
Tangent slope: . At the slope is 1 (45 deg).
Hyperbola with ,
, .
| 0 deg | 3.000 | 0.000 |
| 30 deg | 3.464 | 2.309 |
| 45 deg | 4.243 | 4.000 |
Check at 45 deg: . The parameter must stay away from , where is infinite; the range gives the right branch only.
Answer: parabola points (0,0), (0.75,3), (3,6), (12,12), slope at t = 1 is 1; hyperbola points (3,0), (3.464,2.309), (4.243,4.0).
✓ Answer checked by Claude Opus 5.5 (high effort)
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