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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 y2=4axy^2=4ax: x=at2x=at^2, y=2aty=2at, −∞<t<∞-\infty<t<\infty.
  • Hyperbola x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1: x=asec⁡θx=a\sec\theta, y=btan⁡θy=b\tan\theta (or x=acosh⁡tx=a\cosh t, y=bsinh⁡ty=b\sinh t).

Parabola y2=12xy^2=12x

4a=12⇒a=34a=12\Rightarrow a=3, so x=3t2x=3t^2, y=6ty=6t.

ttx=3t2x=3t^2y=6ty=6t
000
0.50.753
136
21212

Check at t=2t=2: y2=144=12×12y^2=144=12\times12.

Tangent slope: dydx=dy/dtdx/dt=66t=1t\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{6}{6t}=\dfrac1t. At t=1t=1 the slope is 1 (45 deg).

Hyperbola with a=3a=3, b=4b=4

x=3sec⁡θx=3\sec\theta, y=4tan⁡θy=4\tan\theta.

θ\thetaxxyy
0 deg3.0000.000
30 deg3.4642.309
45 deg4.2434.000

Check at 45 deg: 18/9−16/16=2−1=118/9-16/16=2-1=1. The parameter must stay away from θ=±90∘\theta=\pm90^\circ, where sec⁡θ\sec\theta is infinite; the range −90∘<θ<90∘-90^\circ<\theta<90^\circ 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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