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Reciprocal of a Quadratic

recquad.html

Exploration

Given the general quadratic , show that the reciprocal of Q in C is the quadratic

when the auxiliary conic .

Approach

Create a general conic, Q, and the auxiliary conic. Construct a point , assumed to be on Q. Construct the tangent line, L, at . Take the reciprocal of L with respect to C, producing . Show that is on the postulated quadratic.

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 a general quadratic.

Clear[a,b,c,d,e,f];

q1=Quadratic2D[a,b,c,d,e,f];

The point is a point on Q, and L is tangent to Q at .

Clear[x1,y1];

p1=Point2D[x1,y1];

l1=Line2D[p1,q1]

Find the auxiliary conic (a unit circle at the origin).

c1=Circle2D[{0,0},1];

Define the reciprocal function.

Reciprocal2D[

Line2D[A1_,B1_,C1_],

Circle2D[{0,0},1]] :=

Point2D[-A1/C1,-B1/C1];

Find the reciprocal of L.

p2=Reciprocal2D[l1,c1]

Find the reciprocal quadratic.

q2=Quadratic2D[

4*c*f-e^2,2*d*e-4*b*f,

4*a*f-d^2,4*c*d-2*b*e,

4*a*e-2*d*b,4*a*c-b^2];

Construct a polynomial.

eq1=Polynomial2D[q2,Coordinates2D[p2]] //Together

Ignore the denominator and the constant (the numerator will be shown to be zero).

eq2=Numerator[eq1][[2]]

Factor.

eq3=Factor[eq2]

One of the terms is zero, therefore the expression is zero.

eq3 /.

(f+d x1+a x1^2+e y1+b x1 y1+

c y1^2)->0

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