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Computer Aided Geometric Design. Class Exercise #3 Conic Sections – Part 2 5-Point Construction Rational Parameterizations. 1. Q.4. For and , what is the formula for the family of implicit conics passing through and with tangent directions and ?

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computer aided geometric design
Computer Aided Geometric Design

Class Exercise #3

Conic Sections – Part 2

5-Point Construction

Rational Parameterizations

1

slide2
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 .

2

solution
Solution

Recall the following 5-point construction:

3

solution1
Solution

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

4

solution2
Solution

where are the implicit line representations:

5

solution3
Solution

(Check for yourself:

must interpolate !)

6

solution4
Solution

On the other hand:

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

7

solution5
Solution

when and coincide, we have:

8

solution6
Solution

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

9

solution7
Solution

In our case:

10

solution8
Solution

Therefore is:

11

solution9
Solution

The degree of freedom is determined after selecting the point :

hence:

12

solution10
Solution

…. What if ?

13

slide14
Q.5

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

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

14

solution11
Solution

Recall the coordinates given by:

15

solution12
Solution

In this coordinate system:

is an implicit form for :

16

solution13
Solution

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

or, for we write:

17

solution14
Solution

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

18

solution15
Solution

Rearranging:

Therefore the discriminant is:

19

solution16
Solution

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

which is zero for or .

20

solution17
Solution

Going back to the construction we see that gives:

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

21

solution18
Solution

Check yourself:

How can this be?? We constructed with the demand for tangency!

22

solution19
Solution

Hint: What happens to the normal in the degenerate implicit form of ?

Does the analyticdemand of orthogonality with the gradient still mean what we hope it means, geometrically?

23

solution20
Solution

We conclude that in the coordinates the following holds:

Parabola

Ellipse

Hyperbola.

24

solution21
Solution

Some 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 ?

25

slide26
Q.6

Let:

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

26

slide27
Q.6

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

27

slide28
Q.6

Give a rational parameterization of the curve, with

28

solution the quick way
Solution (The Quick Way)

The quick solution:

remember the formula for the special case where :

29

slide30
and for our points this gives:

namely:

and we are done…

Solution (The Quick Way)

30

slide32
Consider the points:

As in the figure:

Solution (The Long Way)

32

slide33
Parameter values of shall correspond to the varying intersection points of the tangent to the curve with the axes:

Choose:

Solution (The Long Way)

33

slide34
and since (special case!):

which gives:

hence:

Solution (The Long Way)

34

slide36
and in lectures the following relation between the tangency point and was shown (by differentiation):

Solution (The Long Way)

36

slide37
Combining gives:

Solution (The Long Way)

37

slide38
By definition of , and from our choice of , we have:

Finally – we can substitute for .

Solution (The Long Way)

38

slide39
Calculate the left term:

Solution (The Long Way)

39

slide40
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)

40

slide42
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

42

slide43
Q.7

Given:

Provide a rational parameterization for the conic section through

Which curve is it?

43

solution22
Solution

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

44

solution23
Solution

We know all but – the product of the ratios.

To find we recall:

so we need to find …

45

solution24
Solution

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

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

46

solution25
Solution

Substituting the forth point in the implicit

equation:

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

47

solution26
Solution

From the usual mapping:

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

48

solution27
Solution

This can be plugged in:

giving:

49

solution28
Solution

Therefore:

This was the last unknown in our parameterization:

50

solution29
Solution

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

51

solution30
Solution

Some 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?

52