Katedra matematiky - Katolícka univerzita v Ružomberku

84 András KovácsIt is easy to calculate that the distance of A and Γ 1 is 4, the distance of Aand Γ 2 is √ 2. By using these equalities and data we get that∣ x−y ∣∣∣ ∣∣√ −|x|2∣ = 4−√ 2.First we make the decomposition of the absolute value in checking with thesoftware.We thought to get 8 sets of points because of the number of the signs ofabsolute values. And really we have got the next equations.Results of the concrete exampleI. x = − √ 2y +6+2 √ 2−y ha y ≤ 2(3+√ 2)√2+2II. x = − √ 2y −6−2 √ 2−y ha y ≤ − 2(3+√ 2)√2+1III. x = √ 2y +10−6 √ 2−y ha y > − 2(3√ 2−5)− √ 2+2IV. x = √ 2y −10+6 √ 2−y ha y ≥ 2(3√ 2−5)√2−1V. x = √ 2y +10−6 √ 2−y ha y < 2(3√ 2−5)√2−1VI. x = − √ 2y +6+2 √ 2−y ha y > 2(3+√ 2)√2+1VII. x = √ 2y −10+6 √ 2−y ha y ≤ 2(3√ 2−5)−2+ √ 2VIII. x = − √ 2y −6−2 √ 2−y ha y > 2(3+√ 2)√2+2So we reached the 3rd hypothesis which means eight rays. We can see thegraphs of these eight rays in the next picture.