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How to find the parabola equation like the picture below, given the vertex $(x$$_o,y_o)$ and theta orientation? please help. thankyou.

tilted parabola

Glorfindel
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cielhive
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    I think you would need more info. To determine the steepness (or shallowness) of the parabola. There are infinite parabolas with the same vertex and theta orientation. – turkeyhundt Nov 30 '14 at 18:12
  • actually, I use it for hough transform in image processing to detect parabola in image. First it use the local gradient information (derivatives dy/dx) to find the Yo coordinates. After that, when I want to find point x,y from xo,yo,theta, I'm confused how to do it. this is the equation to find yo. yo = [ (k1 (x cos θ ο + y sin θ ο) + (x sin θ ο - y cos θ ο)) /(k1sin θ ο - cos θ ο ) ] - xo(k1cos θ ο + sin θ ο) /(k1sin θ ο - cos θ ο ) where k1 = (-sin θο + dy/dx cos θο)/ 2(cos θ ο + dy/dx sin θ ο) – cielhive Nov 30 '14 at 18:21
  • Is that slanted line under angle $\theta$ a directrix of the parabola? – imranfat Nov 30 '14 at 18:32
  • Coordinates x'-y' is the rotation of the coordinates x-y by θ degrees with the centre of the coordinate system as the axis of rotation – cielhive Nov 30 '14 at 18:44

1 Answers1

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Using some trigonometry:

Let $ y = a(x-h)^2 + k $ represent the equation of the parabola at vertex $(h, k)$ on the $xy$ plane

A two dimensional rotation is described as $$x = x'\cos(\theta) + y'\sin(\theta)$$ $$y = -x'\sin(\theta) + y'\cos(\theta)$$ where $(x', y')$ is a point on the $x'y'$plane rotated by $\theta$

Substituting these formulas into the equation of the parabola we get

$$-x'\sin\theta+y'\cos(\theta)=a(x'\cos(\theta) + y'\sin(\theta) - h)^2+k$$

which is probably the most useful form as simplifying gets messy

Here is a desmos link with a rotating parabola: https://www.desmos.com/calculator/hguanwbkbu

dardeshna
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