Katedra matematiky - Katolícka univerzita v Ružomberku

CommentsOn J-conic sections 83We took the Γ ′ 1 , which is parallel to Γ 1 for the examination. (It was thefirst step of suggestion for problem-solving.)The points of the J-hyperbola are at an equal distance from Γ ′ 1 and Γ 2because of the construction of Γ ′ 1 . In this way the points of the J-hyperbolalie on the bisector of Γ ′ 1 and Γ 2.For the details we examined closely this problem. The points of that raywhich starts from G 2 give the bisector of the angle TP ′ PT′′ P, so because ofPTP ′ = PT ′′ ′ P, the next equality is true:PT ′ P −PT ′′P = PT ′ P −PT ′ P ′ = 2a.Similarly, we can get the PTP ′′ − PT′ P= 2a equality, if the points P areon the other ray of e. Consider those points P of e, which can be foundbetween the two rays. These points don’t belong to our solution, becausefor these P, as we can see it easily, the next equality is true:PT ′ P +PT′′ P = 2a.(We were inspired to initiate a new set of points by this new result later.)It is easy to see that another point Q of the plane doesn’t belong to the set ofpoints of theJ-hyperbola. By using the|QTQ ′ −QT′′ Q | = |QT′ Q+2a−QT ′′ ′ Q | =2a equality we get that QTQ ′ = QT ′′ ′ Q. So Q is in the normal bisector of theTQ ′′ T′ Q, which is parallel to the normal bisector of T ′′′ P T′ P. Because of this′fact, the point Q is not an element of the set of the J-hyperbola.A special caseIn that case when Γ 1 is parallel to Γ 2 , the solution is a line, passing throughthe A and parallel to the others.We could see that the process of problem-solving showed some kind of mistake.Since our results, which was taken with geometrical consideration,disagree with the solution of the book, we applied for computers. Becausewe should have liked to try a new learning method, therefore we turn to asoftware which run according to Computer Algebraic System.A concrete exampleWe chose a simple example for checking. The Γ 1 : x = 0, Γ 2 : y = x andA(4;2) simple data lead equations of rays. We∣ know that for point P(x;y)the equalities J(Γ 1 ) = |x| and J(Γ 2 ) =x−y ∣∣∣∣ √ hold.2