Exploring Analyic Geometry with Mathematica®

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Center of a Conic Arc

cacenter.html

Exploration

Show that the center (H,K) of a conic arc whose control points are "cacenter_1.gif", "cacenter_2.gif" and "cacenter_3.gif" and whose projective discriminant is ρ is

"cacenter_4.gif"

"cacenter_5.gif"

where "cacenter_6.gif" is the midpoint of the conic arc's chord and has coordinates

"cacenter_7.gif" and "cacenter_8.gif".

Approach

Form the quadratic equation of a conic arc and convert it to a quadratic. Find the center point of the quadratic and simplify.

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

Determine the quadratic equation of a conic arc.

Clear[a,b,k,x,y,x0,y0,xA,yA,x1,y1,F];
eq1=a*b-k(1-a-b)^2 /.
    {a->((y-yA)(x1-xA)-(x-xA)(y1-yA))/F,
     b->((y-yA)(x0-xA)-(x-xA)(y0-yA))/(-F)} //Simplify

"cacenter_9.gif"

Multiply through by "cacenter_10.gif".

eq2=eq1*F^2

"cacenter_11.gif"

Construct the quadratic and the center points of the quadratic.

q1=Quadratic2D[eq1,{x,y}] //Simplify;
c1=Point2D[q1] //Simplify

"cacenter_12.gif"

Simplify.

Clear[p];
c2=c1 /.
   {F->(y0-yA)(x1-xA)-(x0-xA)(y1-yA),
    k->(1-p)^2/(4p^2),
    x0+x1->2*xM,
    y0+y1->2*yM} //FullSimplify

"cacenter_13.gif"

Change the signs on the numerator and denominator to get the desired formulas.

Map[((-1*Numerator[#])/(-1*Denominator[#]))&,c2]

"cacenter_14.gif"


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