Exploring Analyic Geometry with Mathematica®

Home Contents Commands Packages Explorations Reference
Tour Lines Circles Conics Analysis Tangents

Conic Arcs

Definition of a Conic Arc
Equation of a Conic Arc
Projective Discriminant
Conic Characteristics
Parametric Equations
Explorations

In previous chapters we introduced line segments and circular arcs which are pieces of more complete curves. In this chapter we introduce a conic arc which is a piece of a conic curve. As with circular arcs, conic arcs are useful for constructing smoothly connected sequences of curves as well as pleasing aesthetic shapes.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Definition of a Conic Arc [Top]

"conarc_1.gif"

Definition of a conic arc.

Let points "conarc_2.gif" and "conarc_3.gif" be the start and end points, respectively, of a segment of a conic curve, Q, and let "conarc_4.gif" be the point of intersection, or apex, of the two tangent lines to the curve at "conarc_5.gif" and "conarc_6.gif" as shown in Figure [conarc:fig01]. Furthermore, let h equal the maximum height of the segment measured from the chord "conarc_7.gif" and k be the distance from "conarc_8.gif" to the chord "conarc_9.gif". The points "conarc_10.gif", "conarc_11.gif", "conarc_12.gif" and the ratio ρ, given by ρ=h/k, define a conic arc. The ratio ρ is called the projective discriminant of the conic arc, and the point at the maximum height on the curve is called the shoulder point. The points "conarc_13.gif", "conarc_14.gif" and "conarc_15.gif" are called control points of the conic arc.

Example: Plot the conic arc with start and end points (-2,1) and (3,0), respectively, apex point (1,2) and projective discriminant ρ=0.45.

Solution: Descarta2D represents a conic arc as

ConicArc2D[{x"conarc_16.gif",y"conarc_17.gif"},{x"conarc_18.gif",y"conarc_19.gif"},{x"conarc_20.gif",y"conarc_21.gif"},ρ]

where "conarc_22.gif" and "conarc_23.gif" are the coordinates of the start and end points, respectively, "conarc_24.gif" is the apex point and ρ is the projective discriminant.

Sketch2D[{c1=ConicArc2D[{-2,1},{1,2},{3,0},0.45]},
         PlotRange->{{-3,3},{-1,2}}]

"conarc_25.gif"

There are several functions provided by Descarta2D to query conic arcs. The function Rho2D[cnarc] returns the ρ value of the conic arc. Point2D[cnarc,Apex2D] returns the apex control point of a conic arc. The coordinates of points on a conic arc at a parameter value t are returned by the function cnarc[t], t=0 returns the start point coordinates, t=1 the end point coordinates and t=1/2 the shoulder point coordinates.

{Rho2D[c1],
Point2D[c1,Apex2D],
Map[c1[#]&, {0, 1/2, 1}]}

"conarc_26.gif"

Equation of a Conic Arc [Top]

The curve underlying the conic arc is clearly a conic curve since there are five conditions imposed on the curve (two points, two tangents and the projective discriminant, ρ). The projective discriminant, ρ, can be interpreted as defining a third line tangent to the curve, parallel to the line "conarc_27.gif" at a distance h from "conarc_28.gif", where h is given by h=ρk and k is the distance from "conarc_29.gif" to line "conarc_30.gif".

In a subsequent chapter we will describe a general procedure for finding the quadratic equation of a conic constrained by two points and three tangent lines, and we will show that when the two points are on two of the tangent lines, there is only one quadratic satisfying the conditions. Specifically, the equation is given by

"conarc_31.gif"

where,

k = "conarc_32.gif"
α = "conarc_33.gif"
β = "conarc_34.gif"

Example: Find the quadratic associated with the conic arc with start and end points (0,0) and (3,0), respectively, apex point (1,2) and projective discriminant ρ=1/4. Find the conic curve associated with the conic arc.

Solution: The function Quadratic2D[cnarc] returns the quadratic associated with a conic arc. The function Loci2D[cnarc] returns a list containing the conic curve associated with a conic arc.

ca1=ConicArc2D[{0,0},{1,2},{3,0},1/4];
{q1=Quadratic2D[ca1] //Simplify,
c1=Loci2D[ca1] //N}

"conarc_35.gif"

Map[Sketch2D[{#},PlotRange->{{-1,4},{-2,1}}]&,
    {ca1,c1}]

"conarc_36.gif"

Projective Discriminant [Top]

In this section we will examine the significance of the value of the projective discriminant, ρ. By definition, ρ may take on values in the range

0<ρ<1.

Consider the conic arc, S, with start and end points (0,0) and (1,0), respectively, apex point "conarc_37.gif" and projective discriminant ρ. Clearly, any arbitrary conic arc can be transformed to coincide with S by applying a proper sequence of translations, rotations and scaling transformations. Such transformations do not change the type of conic curve associated with the conic arc. Using Mathematica we can find the quadratic equation underlying S as shown by the following commands.

Clear[xA,yA,p];
S=ConicArc2D[{0,0},{xA,yA},{1,0},p];
Q=Quadratic2D[S] //Simplify

"conarc_38.gif"

As has already been shown in a previous chapter, the specific conic type of the quadratic equation

"conarc_39.gif"

is determined by the discriminant, "conarc_40.gif". For an ellipse D<0, for a parabola D=0, and for a hyperbola D>0. For the quadratic, Q, representing the conic arc S defined above, "conarc_41.gif". It is clear by inspection, that if 0<ρ<1/2 the conic is an ellipse; if ρ=1/2 the conic is a parabola; and for 1/2<ρ<1 the conic is a hyperbola.

Conic Characteristics [Top]

In Section [conarc:sec01] we showed that the quadratic equation associated with a conic arc is given by

"conarc_42.gif"

where,

k = "conarc_43.gif"
α = "conarc_44.gif"
β = "conarc_45.gif"

Therefore, since we know its quadratic equation, all the geometric characteristics of the conic curve associated with the conic arc can be expressed in terms of the defining elements of the conic arc, "conarc_46.gif", "conarc_47.gif", "conarc_48.gif" and ρ. Of particular interest is the formula for the center of a central conic (circle, ellipse or hyperbola), since this formula is used in the next section to convert a conic into a conic arc. The center point (H,K) is given by

H = "conarc_49.gif"
K = "conarc_50.gif"

where "conarc_51.gif" is the midpoint of the conic arc's chord and has coordinates

"conarc_52.gif"

This formula is derived in the exploration cacenter.html.

Example: Find the center of the conic arc with control points (0,0), (2,1) and (3,0) and ρ=1/4.

Solution: The Descarta2D function Point2D[cnarc] returns the center point of a conic arc (the underlying conic cannot be a parabola).

ca1=ConicArc2D[{0,0},{2,1},{3,0},1/4];
Point2D[ca1]

"conarc_53.gif"

Let Q be a conic and L be a line that intersects Q in two distinct points. We wish to determine the conic arc, S cut by L through Q. Clearly, the intersection points of the line L with Q are the start and end points of S. Also, the line passing through the intersection points is the polar (line) of the apex point, "conarc_54.gif", of S. To complete the definition of the conic arc, we need to determine ρ. If the conic is a parabola, then ρ=1/2; otherwise, we can assume that the conic is a central conic. Assume the center of the conic is (h,k). Then, using the formula for the x-coordinate of the center of a conic arc given in Equation ([conarc:eqn01]), we solve to find the value of ρ to be

"conarc_55.gif"

where "conarc_56.gif" is the midpoint of "conarc_57.gif". We choose the plus sign in the denominator because ρ has to be less than one and the radical produces a positive number. In certain configurations, this formula will be indeterminate and we instead use the y-coordinate of the center of the conic arc yielding

"conarc_58.gif"

again choosing the plus sign in the denominator.

Example: Find the conic arc cut by the line 2x-4y=0 through the ellipse

"conarc_59.gif"

Plot the original curves and the conic arc separately.

Solution: The Descarta2D function ConicArc2D[line,conic] returns a conic arc defined by a line cutting a conic curve.

l1=Line2D[2,-4,0];
e1=Ellipse2D[{1,-1},3,2,0];
ca1=ConicArc2D[l1,e1]

"conarc_60.gif"

Map[Print[Sketch2D[{#},PlotRange->{{-5,5},{-3,3}}]]&,
    {{l1,e1},ca1}];

"conarc_61.gif"

"conarc_62.gif"

Parametric Equations [Top]

The conic arc defined in this chapter is a special case of a more general curve called a rational quadratic Bézier. The parametric equations of this simplified formulation are given by

x = "conarc_63.gif"
y = "conarc_64.gif"

where ρ=h/k is the projective discriminant and

"conarc_65.gif" = "conarc_66.gif"
"conarc_67.gif" = 2t(1-t)
"conarc_68.gif" = "conarc_69.gif"

It is clear from direct substitution that "conarc_70.gif" is the point whose coordinates correspond to t=0, and "conarc_71.gif" corresponds to t=1. The point where the curve intersects the line through the midpoint of "conarc_72.gif" and "conarc_73.gif" is called the shoulder point of the conic arc. The shoulder point corresponds to the parameter value t=1/2.

Example: Plot nine points at equal parameter values on the conic arc with (-2,1) and (1,2) as start and end point, (0,3) as the apex point and ρ=0.45.

Solution: The Descarta2D function cnarc[t] returns the coordinates of a point on a conic arc at a parameter t.

ca1=ConicArc2D[{-2,1},{0,3},{1,2},0.45];
Sketch2D[{ca1,Map[Point2D[ca1[#]]&,Range[0,8]/8]}]

"conarc_74.gif"

Explorations [Top]

Circular Conic Arc. [cacircle.html]

Show that the conic arc with control points (0,0), (a,b) and (2a,0) will be a circular arc if

"conarc_75.gif"

Center of a Conic Arc. [cacenter.html]

Show that the center point (H,K) of a conic arc whose control points are "conarc_76.gif", "conarc_77.gif" and "conarc_78.gif" and projective discriminant ρ is

H = "conarc_79.gif"
K = "conarc_80.gif"

where "conarc_81.gif" is the midpoint of the conic arc's chord and has coordinates

"conarc_82.gif"

Tangent Line at Shoulder Point. [catnln.html]

Let P be the point at parameter value t=1/2 on a unit conic arc, C, whose control points are "conarc_83.gif", "conarc_84.gif" and "conarc_85.gif" and whose projective discriminant is ρ. Let L be the line tangent to C at t. Show that L is parallel to the chord "conarc_86.gif" at a distance from "conarc_87.gif". The point P is called the shoulder point of the conic arc.

Coordinates of Shoulder Point. [shoulder.html]

Show that the coordinates of the shoulder point of a conic arc with control points "conarc_88.gif", "conarc_89.gif" and "conarc_90.gif" and projective discriminant ρ are given by

"conarc_91.gif"

where "conarc_92.gif" is the midpoint of the conic arc's chord and has coordinates

"conarc_93.gif"

Shoulder Point on Median. [camedian.html]

Let C be a conic arc with control points "conarc_94.gif", "conarc_95.gif" and "conarc_96.gif" and projective discriminant ρ. Let P be the point on the median "conarc_97.gif" associated with vertex "conarc_98.gif" of triangle "conarc_99.gif" such that "conarc_100.gif" ("conarc_101.gif" is the midpoint of "conarc_102.gif"). Show that P is coincident with the shoulder point of C, having coordinates

"conarc_103.gif"

Parametric Equations of a Conic Arc. [caparam.html]

Show that the parametric equations of a unit conic arc represent the same implicit quadratic equation as the one underlying the conic as derived from the control points "conarc_104.gif", "conarc_105.gif" and "conarc_106.gif" and ρ.


Copyright © 1999-2007 Donald L. Vossler, Descarta2D Publishing
www.Descarta2D.com