Katedra matematiky - Katolícka univerzita v Ružomberku

86 András KovácsPT ′ P +PT′′ P = AT′ A +AT′′ A = 2aequality. The figure of this set of points is the segment G 1 G 2 , as we seebelow. It is easy to see that this segment lies on the bisector of Γ ′ 1 and Γ′ 2 .(We made mention of this fact in connection with figure 1.)Figure 3. The J-ellipse belonging to two linesThe analysis of the J-ellipseThe J-ellipse is on the bisector of the TA ′′ G 2TA ′ , since T ′ ′ AAG ′ 2 and TA ′′ AG 2are congruent triangles. So if P is a point of the segment, then the nextequality is true:PT ′ P +PT′′ P = PT′ P +PT′ P ′ = 2aIf Γ 1 is parallel to Γ 2 , then the solution - instead of the segment G 1 G 2 - isa line which passing through A and parallel to Γ 1 and Γ 2 .With the help of the previously givenΓ 1 ,Γ 2 andA, we got the next equation: