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Computer Aided Geometric Design

Class Exercise #3

Conic Sections – Part 25-Point ConstructionRational Parameterizations

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Q.4

For and , what is the formula for the family of implicit conics passing through and with tangent directions and ?

Calculate the value of the last degree of freedom which could force the conic to interpolate through an arbitrary point .

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Solution

Recall the following 5-point construction:

𝑃1

𝑃2 𝑃3

𝑃4

𝑃5

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SolutionWhen seeking a conic that interpolates , we write the one parameter family of implicit conics:

𝑃3

𝑃4

𝑃1

𝑃2𝑃5

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Solutionwhere are the implicit line representations:

𝑃3

𝑃4

𝑃1

𝑃2𝑃5

𝐿1𝐿2𝐿3

𝐿4

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Solution(Check for yourself:

must interpolate !)

𝑃3

𝑃4

𝑃1

𝑃2𝑃5

𝐿1𝐿2𝐿3

𝐿4

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SolutionOn the other hand:

must be a conic section, since it is a quadratic form.

𝑃3

𝑃4

𝑃1

𝑃2𝑃5

𝐿1𝐿2𝐿3

𝐿4

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Solutionwhen and coincide, we have:

𝑃3

𝑃4

𝑃1

𝑃2𝑃5

𝐿1𝐿2

𝐿3𝐿4

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SolutionWhich means we require tangency to and and interpolation of three points.

𝑃3=𝑃4𝑃1=𝑃2

𝑃5

𝐿1𝐿2

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SolutionIn our case:

𝑃2=(1 ,2)

𝑃1=(0 ,0)

𝑄

𝐿1𝐿2

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SolutionTherefore is:

𝑃2=(1 ,2)

𝑃1=(0 ,0)

𝑄𝐿1𝐿2

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SolutionThe degree of freedom is determined after selecting the point :

hence:

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Solution

…. What if ?

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Q.5

1. Given three points and , and a blending constant , determine the discriminant of the resulting conic.

2. For which values of is it an ellipse? A parabola? A hyperbola?

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SolutionRecall the coordinates given by:

𝑃1

𝑃2𝑇

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SolutionIn this coordinate system:

is an implicit form for : 𝑃1

𝑃2𝑇

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SolutionTherefore, by the same construction as in the previous question, the implicit curve is given by:

or, for we write:

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SolutionWe now find the discriminant with respect to the implicit conic in the coordinates:

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SolutionRearranging:

Therefore the discriminant is:

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Solutionwe see that this is a quadratic expression in . What are the possible signs?

which is zero for or .

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SolutionGoing back to the construction we see that gives:

This is a singular case that does not satisfy our geometric demands of tangency at and .

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Solution

Check yourself:How can this be?? We constructed with the

demand for tangency !

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Solution

Hint: What happens to the normal in the degenerate implicit form of ? Does the analytic demand of orthogonality with the gradient still mean what we hope it means, geometrically?

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SolutionWe conclude that in the coordinates the following holds:

Parabola Ellipse

Hyperbola.

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SolutionSome points to think about:

How, if at all, does the answer change in the coordinate system?

How is the above related to the coordinate change mapping ?

Why were the values of restricted? How is this related to the values of ?

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Q.6

Let:

Assume , where and are the ratios determined by the tangent to a point on the curve (as in the next slide).

Recall from lectures that is constant for the entire curve (in fact characterizing it)

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Q.6

𝑃2

𝑇𝑃1

𝐴2

𝐴1

Give a rational parameterization of the curve, with

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Q.6

𝑃2

𝑇𝑃1

𝐴2

𝐴1

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Solution (The Quick Way)The quick solution:remember the formula for the special case where :

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and for our points this gives:

namely:

and we are done…

Solution (The Quick Way)

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Let’s do the long way once to review the geometry behind the formula.

Solution (The Long Way)

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Consider the points:

As in the figure:

Solution (The Long Way)

𝑃2

𝑇𝑃1

𝐴2

𝐴1

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Parameter values of shall correspond to the varying intersection points of the tangent to the curve with the axes:

Choose:

Solution (The Long Way)

𝑃2

𝑇𝑃1

𝐴2

𝐴1

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and since (special case!):

which gives:

hence:

Solution (The Long Way)

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Now, each point on our curve is of the form:

Solution (The Long Way)

𝑃2

𝑇𝑃1

𝐴2

𝐴1

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and in lectures the following relation between the tangency point and was shown (by differentiation):

Solution (The Long Way)

𝑃2

𝑇𝑃1

𝐴2

𝐴1

𝑃𝑐=(𝑢𝑐 ,𝑣𝑐)

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Combining gives:

Solution (The Long Way)

𝑃2

𝑇𝑃1

𝐴2

𝐴1

𝑃𝑐=(𝑢𝑐 ,𝑣𝑐)

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By definition of , and from our choice of , we have:

Finally – we can substitute for .

Solution (The Long Way)

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Calculate the left term:

Solution (The Long Way)

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A similar calculation gives for the right term, which means we got the same result as in the short solution with the formula.

Solution (The Long Way)

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We constructed a quadratic parametric form:

A Final Remark

𝑃2

𝑇𝑃1

𝐴2

𝐴1

𝑃𝑐=(𝑢𝑐 ,𝑣𝑐)

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which is a combination of three given points on a polygon with special blending coefficients.

This is a special case of much more general representations you shall see later in lectures.

A Final Remark

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Q.7

Given:

Provide a rational parameterization for the conic section through Which curve is it?

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SolutionWe shall parameterize with . The following was shown in lectures:

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SolutionWe know all but – the product of the ratios.

To find we recall:

so we need to find …

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SolutionRemember that are all equivalent ways of specifying the unique conic.

They are determined, in our case, by the fourth given point:

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SolutionSubstituting the forth point in the implicit equation:

is not yet possible, since it is given in coordinates.

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SolutionFrom the usual mapping:

This is an easy linear system in as unknowns, solution of which is:

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SolutionThis can be plugged in:

giving:

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SolutionTherefore:

This was the last unknown in our parameterization:

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Solution

This is a hyperbola, since , which is smaller than .

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SolutionSome points to think about:

Again - Why did we conclude a hyperbola from the value of with respect to the system?

What part of the curve did we get? The entire hyperbola? Part of it?

What is the meaning of the values where the denominator is zero?