Computer Aided Geometric Design

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 ? Calculate the value of the last degree of freedom which could force the conic to interpolate through an arbitrary point . 2

Solution Recall the following 5-point construction: 3

Solution When seeking a conic that interpolates , we write the one parameter family of implicit conics: 4

Solution where are the implicit line representations: 5

Solution (Check for yourself: must interpolate !) 6

Solution On the other hand: must be a conic section, since it is a quadratic form. 7

Solution when and coincide, we have: 8

Solution Which means we require tangency to and and interpolation of three points. 9

Solution In our case: 10

Solution Therefore is: 11

Solution The degree of freedom is determined after selecting the point : hence: 12

Solution …. What if ? 13

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

Solution Recall the coordinates given by: 15

Solution In this coordinate system: is an implicit form for : 16

Solution Therefore, by the same construction as in the previous question, the implicit curve is given by: or, for we write: 17

Solution We now find the discriminant with respect to the implicit conic in the coordinates: 18

Solution Rearranging: Therefore the discriminant is: 19

Solution we see that this is a quadratic expression in . What are the possible signs? which is zero for or . 20

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

Solution Check yourself: How can this be?? We constructed with the demand for tangency! 22

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

Solution We conclude that in the coordinates the following holds: Parabola Ellipse Hyperbola. 24

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

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

Q.6 Recall from lectures that is constant for the entire curve (in fact characterizing it) 27

Q.6 Give a rational parameterization of the curve, with 28

Solution (The Quick Way) The quick solution: remember the formula for the special case where : 29

and for our points this gives: namely: and we are done… Solution (The Quick Way) 30

Let’s do the long way once to review the geometry behind the formula. Solution (The Long Way) 31

Consider the points: As in the figure: Solution (The Long Way) 32

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

and since (special case!): which gives: hence: Solution (The Long Way) 34

Now, each point on our curve is of the form: Solution (The Long Way) 35

and in lectures the following relation between the tangency point and was shown (by differentiation): Solution (The Long Way) 36

Combining gives: Solution (The Long Way) 37

By definition of , and from our choice of , we have: Finally – we can substitute for . Solution (The Long Way) 38

Calculate the left term: Solution (The Long Way) 39

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

We constructed a quadratic parametric form: A Final Remark 41

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

Q.7 Given: Provide a rational parameterization for the conic section through Which curve is it? 43

Solution We shall parameterize with . The following was shown in lectures: 44

Solution We know all but – the product of the ratios. To find we recall: so we need to find … 45

Solution Remember that are all equivalent ways of specifying the unique conic. They are determined, in our case, by the fourth given point: 46

Solution Substituting the forth point in the implicit equation: is not yet possible, since it is given in coordinates. 47

Solution From the usual mapping: This is an easy linear system in as unknowns, solution of which is: 48

Solution This can be plugged in: giving: 49

Solution Therefore: This was the last unknown in our parameterization: 50

Computer Aided Geometric Design