To construct a conic (c) knowing two tangents of it, its contact point with one of them and one of its foci.

The conic can be easily constructed using the property in DirectrixProperty.html . In fact, let the known tangents t_{C}, t_{X}
intersect at point B and C be the known contact point on one of them.Then, if X is the contact point on the other
tangent angle(XAB) = angle(BAC) and this determines X as intersection of the known tangent with the reflected of
AC on AB. The other focus is then at a point Y such that angle AXY is bisected by XB. Analogously Y is on line CY
such that angle ACY is bisected by CB. This allows the construction of Y as intersection point of two known lines. Having the two foci {A,Y} and points {C,X} on the conic later is easily constructed.

Remark-1 It is easy also to find the kind of the conic from the given data.In fact, let U be the intersection of XC with line AB. Then the harmonic
conjugate V=U(X,C) of U w.r. to (X,C) is on the polar of A i.e. the directrix corresponding to A.Thus, drawing from V an orthogonal line to
AY we construct the directrxix d_{A}.From this we can determine the eccentricity of the conic by measuring the ratio of distances of X from the
focus A and the directrix d_{A}. Remark-2 The case in which Y is at infinity i.e. lines XY and CY are parallel corresponds to the parabola.