---
title: "Parametric parabola and hyperbola points"
subject: "Computer Aided Design"
chapter: 3
key: "semester_4/computer_aided_design:conics-parabola-hyperbola"
source: https://www.bctengineer.com/q/semester_4/computer_aided_design/conics-parabola-hyperbola
site: BCT Engineering
---

# Parametric parabola and hyperbola points

Computer Aided Design · Chapter 3: Mathematical Representation of Curves

- Asked in: year not given, 6 marks
- Video: [Parametric Equation of Parabola and Hyperbola - Techniques For Geometric Modeling - CAD/CAM/CAE](https://www.youtube.com/watch?v=-oYuPLfB7Hs) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/computer_aided_design/conics-parabola-hyperbola
- Whole chapter: https://www.bctengineer.com/md/semester_4/computer_aided_design/chapter-3

## Question

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.

## 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).

## Answer

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 $y^2=4ax$: $x=at^2$, $y=2at$, $-\infty<t<\infty$.
- Hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$: $x=a\sec\theta$, $y=b\tan\theta$ (or $x=a\cosh t$, $y=b\sinh t$).

#### Parabola $y^2=12x$
$4a=12\Rightarrow a=3$, so $x=3t^2$, $y=6t$.

| $t$ | $x=3t^2$ | $y=6t$ |
|---|---|---|
| 0 | 0 | 0 |
| 0.5 | 0.75 | 3 |
| 1 | 3 | 6 |
| 2 | 12 | 12 |

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

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

#### Hyperbola with $a=3$, $b=4$
$x=3\sec\theta$, $y=4\tan\theta$.

| $\theta$ | $x$ | $y$ |
|---|---|---|
| 0 deg | 3.000 | 0.000 |
| 30 deg | 3.464 | 2.309 |
| 45 deg | 4.243 | 4.000 |

Check at 45 deg: $18/9-16/16=2-1=1$. The parameter must stay away from $\theta=\pm90^\circ$, where $\sec\theta$ is infinite; the range $-90^\circ<\theta<90^\circ$ 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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For y^2 = 12x, what is a?
   A. 12
   B. 6
   C. 4
   D. 3
   Answer: D. 4a = 12, so a = 3, giving x = 3t^2, y = 6t. (Key point: Parabola parametric)

2. What is the slope of the tangent to the parabola at t = 1?
   A. 1
   B. 6
   C. 0.5
   D. 1/6
   Answer: A. dy/dx = (dy/dt)/(dx/dt) = 6/(6t) = 1/t, which is 1 at t = 1. (Key point: Tangent slope)

3. Which parametric form does the answer give for x^2/a^2 - y^2/b^2 = 1?
   A. x = a cos θ, y = b sin θ
   B. x = a sec θ, y = b tan θ
   C. x = a tan θ, y = b sec θ
   D. x = a sin θ, y = b cos θ
   Answer: B. x = a sec θ, y = b tan θ gives the right branch, with θ kept away from 90°. (Key point: Hyperbola parametric)

4. What is the parabola point at t = 2?
   A. (6, 12)
   B. (12, 6)
   C. (12, 12)
   D. (4, 12)
   Answer: C. x = 3(4)=12, y = 6(2)=12. (Key point: Parabola point)

## Related in Computer Aided Design

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