Exploring Analyic Geometry with Mathematica®

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Tangent Line at Shoulder Point

catnln.html

Exploration

Let P be the point at parameter value t=1/2 on a unit conic arc, C, whose control points are "catnln_1.gif", "catnln_2.gif" and "catnln_3.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 "catnln_4.gif" at a distance b ρ from "catnln_5.gif". The point P is called the shoulder point of the conic arc.

Approach

Create the conic arc and construct a point at t=1/2. Construct the quadratic underlying the conic arc. Construct the polar of P with respect to the quadratic (the tangent, L). Show that L is horizontal and, therefore, parallel to the conic arc's chord.

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`

Solution

Create the conic arc.

Clear[a,b,p];
ca1=ConicArc2D[{0,0},{a,b},{1,0},p];

Construct the point at t=1/2.

P=Point2D[ca1[1/2]] //Simplify

"catnln_6.gif"

Construct the underlying quadratic.

Q=Quadratic2D[ca1] //Simplify

"catnln_7.gif"

The tangent line at P is horizontal and at a distance b ρ from "catnln_8.gif".

L=Line2D[P,Q] //Simplify

"catnln_9.gif"

Discussion

Plot a numerical example with a=1, b=2 and ρ=0.45.

Sketch2D[{ca1,P,L} /. {a->1,b->2,p->0.45},
   CurveLength2D->5, PlotRange->{{-0.25,1.25},{-0.25,1.25}}]

"catnln_10.gif"

Graphics saved as "catnln01.eps".


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