A preliminary discussion on the set of conics inscribed in a given quadrilateral may be found in CircumscriptibleQuadrilateral2.html . The solution uses the fact that the centers of all these inscribed conics are on the

This is a special case of the fact that the poles of a given line with respect to all inscribed conics are on another line. Taking for given line the line at infinity we conclude that the centers of the inscribed conics (which are the poles of the line at infinity) are on a line. By specializing to the (degenerate) tangent conics coinciding with the end-points of the diagonal segments we identify this line with the Newton line as stated. The center in the case of the parabola coincides with the point at infinity of this line, which thus determines also the direction of the axis of the parabola.

To find the parabola stick to a diagonal point of the quadrilateral, G say, and draw from there a parallel (n) to the Newton line. Then find the conjugate (n') with respect to the sides {AD,BC} through G. From the intersection point E of the diagonals draw a parallel EL' to (n') and its conjugate EM' with respect to the diagonals {EF,EJ}. Line EL' intersects sides {AD,BC} at two points {K,L} which are contact points of the parabola. The same is true for the intersection points {M,N} of line EM' with the other two sides of the quadrilateral.

For the actual construction I consider the bitangent family of conics tangent to lines {AD,BC} at points {K,L}. The parabola is the member passing through one of the other points {M,N}.

The clue idea is that the parallel to the Newton line through G passes through the middle S of KL, hence later is parallel to the conjugate of GS with respect to {AD,BC}. This determines the direction of EL' (and its conjugate EM' with respect to {EF,EJ}) necessary for the determination of the contact points of the inscribed conic.

CircumscriptibleQuadrilateral.html

CircumscriptibleQuadrilateral2.html

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