## Exploring Analyic Geometry with |
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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 , and and whose projective discriminant is ρ is

where is the midpoint of the conic arc's chord and has coordinates

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

Multiply through by .

eq2=eq1*F^2

Construct the quadratic and the center points of the quadratic.

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

c1=Point2D[q1] //Simplify

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

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

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

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