КАНОНИЧЕСКИЕ УРАВНЕНИЯ КРИВЫХ ВТОРОГО ПОРЯДКА

Аннотация

This work covers one of the important topics of the analytical geometry section, the canonical equations of second-order curves. Based on the general equation of curves, their classification, geometric properties, and methods of simplifying the equation by rotating the coordinate axis are analyzed. Also, the canonical equations of second-order curves such as ellipse, hyperbola, and parabola, their graph construction, center, and symmetry properties are widely covered. Keywords: Second-order equation, curves, canonical equation, ellipse, hyperbola, parabola. Language: Uzbek Citation: Azimov , R. (2025). CANONICAL EQUATIONS OF SECOND-ORDER CURVES. Universal International Scientific Journal, 2(1), 192–201. Retrieved from https://universaljurnal.uz/index.php/jurnal/article/view/1492 Doi: https://doi.org/10.5281/zenodo.15227510 Google scholar: https://scholar.google.com/scholar?hl=ru&as_sdt=0%2C5&q=IKKINCHI+TARTIBLI+EGRI+CHIZIQLARNING+KONONIK+TENGLAMALARI&btnG= √(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = R Kirish. Ushbu maqolada ikkinchi yoki tartibli egri chiziqlarning kanonik tenglamalari yoritiladi. Ikkinchi tartibli (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑅2. egri chiziqlardan aylana, ellips, ellipsning (1.1) parametrik tenglamasi, giperbola, Aylana tenglamalari quyidagi parabolalarning ta’rifi, tenglamalari va ko‘rinishlarda bo‘lishi mumkin: grafik tasvirlari misollar yordamida keltirib o‘tiladi. (𝑥 − 𝑎)2 + 𝑦2 = 𝑅2, S (a;0), R 0 Aylana. Ta’rif. Tekislikda berilgan 𝑥2 + (𝑦 − 𝑏)2 = 𝑅2, S (0;b), R nuqtadan bir xil uzoqlikda joylashgan 0 nuqtalarning geometrik o‘rniga aylana 𝑥2 + 𝑦2 = 𝑅2, S (0;0), R 0 deyiladi. Agar M (x ; y ), nuqta aylana tashqarisida 1 1 1 Faraz qilaylik S (a,b) – aylana 0 bo‘lsa, (x − 𝑎)2 + (y + b)2 > 𝑅2 1 1 markazi bo‘lsin. Aylanani ixtiyoriy tengsizlik o‘rinli bo‘ladi. nuqtasidan markazgacha bo‘lgan masofani Agar M (x ; y ) nuqta aylana ichida 2 2 2 R harfi bilan belgilaylik (1-rasm). joylashgan bo‘lsa, (x − 𝑎)2+ (y + b)2 < Faraz qilaylik M(x,y) – aylananing 2 2 𝑅2 tengsizlik o‘rinli bo‘ladi. ixtiyoriy nuqtasi bilan. Ta’rifga ko‘ra |S 𝑀| = 𝑅. Ikki nuqta orasidagi masofa 0 formulasiga ko‘ra |S 𝑀| = √(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 0 Bundan esa , 193 Universaljurnal.uz Universal International Scientific Journal 2025, 2(1) (𝑥 − 4)2 + (𝑦 + 3)2 = 32 Bu tenglamadan ko‘rinadiki berilgan chiziq aylana ekan, uning markazi S (4;-3) nuqtada, radiusi esa R = 0 3. Ellips. Ta’rif. Tekkislik nuqtalaridan focus deb ataluvchi 𝐹 va 𝐹 nuqtalarga 1 2 bo‘lgan masofalarni yig‘indisi o‘zgarmas 1-rasm. bo‘lgan nuqtalarning geometrik o‘rniga Quyidagi tenglamani qaraylik: ellips deyiladi. Faraz qilaylik 𝐹 va 𝐹 lar ellipsning 𝐴𝑥2 + 𝐴𝑦2 + 𝐷𝑥 +Ey + F = 0 1 2 fokuslari bo‘lsin va ular orasidagi masofani (1.2) 2C bilan belgilaylik: (2-rasm). bu erda 𝐴 ≠ 0. Bu tenglamani barcha |F F | = 2C 1 2 xadlarini A ga bo‘lib quyidagi tenglamani hosil qilamiz: 𝐷 𝐸 𝐹 𝑥2 + 𝑦2 + x + y + = 0 A A A (1.3) yoki 𝐷 𝐸 𝐷2 + 𝐸2−4AF (𝑥 + ) 2 + (𝑦 + ) 2 = 2A 2A 4𝐴2 (1.4) 2-rasm. Demak bundan ko‘rinadiki yuqori darajalar oldidagi koeffitsintlar bir xil son bo‘lsa, bu Ikki nuqta orasidagi masofa formulasidan chiziq aylana bo‘lishi mumkin ekan. foydalanib |F 𝑀| va |F 𝑀| larni topamiz. 1 2 1-Misol. Quyidagi tenglama bilan berilgan aylanani markazi va radiusi |F 𝑀| = √(𝑥 + 𝑐)2 + 𝑦2 , |F 𝑀| = 1 2 topilsin: √(𝑥 − 𝑐)2 + 𝑦2 𝑥2 + 𝑦2 – 8x + 6y +16 = 0 Ta’rifga ko‘ra Yechish: Bu tenglamani quyidagi ko‘rinishda yechish ham mumkin: √(𝑥 + 𝑐)2 + 𝑦2 + √(𝑥 − 𝑐)2 + 𝑦2 = 2a (2.1) (𝑥2 − 8𝑥 + 16) + (𝑦2 + 6𝑦 + 9) − 9 + 16 − 10 = 0 Hosil bo‘lgan ellipsni tenglamasini soddaroq holda keltirish uchun uni yoki quyidagi ko‘rinishda yozamiz 194 Universaljurnal.uz Universal International Scientific Journal 2025, 2(1) √(𝑥 − 𝑐)2 + 𝑦2 = 2a−√(𝑥 − 𝑐)2 + 𝑦2 𝑥2 + 𝑦2 = 1 dan 𝑦2 = 𝑏2 (𝑎2 − 𝑥2) bu 𝑎2 𝑏2 𝑎2 ifodani yuqoridagi 𝑦2 ni o‘rniga olib borib Tenglikni har ikki tomonini kvadratga qo‘yamiz. ko‘taramiz. 𝑥2 + 2cx + 𝑐2 + 𝑦2 = 4𝑎2 – 𝑏2 r = √𝑥2 − 2cx + 𝑐2 + (𝑎2 − 𝑥2) = 4a√(𝑥 − 𝑐)2 + 𝑦2 + 𝑥2 -2cx + 𝑐2 + 𝑦2. 2 𝑎2 𝑥2𝑎2 −2cx𝑎2 +𝑎2 𝑐2+𝑏2𝑎2−𝑏2𝑥2 Bu yerdan quyidagi ifoda kelib chiqadi: 𝑎2 𝑎√(𝑥 − 𝑐)2 + 𝑦2 = 𝑎2 – cx 𝑥2(𝑎2 −𝑏2)−2cx𝑎2 +𝑎2(𝑐2+𝑏2) = √ = 𝑎2 Hosil bo‘lgan ifodani esa kvadratga 𝑥2𝑐2−2cx𝑎2 +𝑎4 (cx−𝑎2)2 √ = √ ko‘taramiz. 𝑎2 𝑎2 𝑥2(𝑎2 − 𝑐2) + 𝑦2𝑎2 = 𝑎2(𝑎2 − 𝑐2) 𝑐 𝑥 − 𝑎 2 c = = 𝑥 − 𝑎, bu yerda 𝑎2 − 𝑐2 = 𝑏2 (2.2) a a c ekanligidan foydalandik. = e belgilaymiz a Quyidagi belgilashni kiritamiz: va bu sonni ellipsning ekssentrisiteti deb ataymiz. Natijada r fakal radius uchun 𝑎2 − 𝑐2 = 𝑏2 2 quyidagi formulani hosil qilamiz r = (2.3) 2 𝑒𝑥 − 𝑎. Natijada quyidagi tenglamaga ega 𝑟 fakal radius uchun r = a – yex ifodani bo‘lamiz: 1 1 hosil qilish mumkin. 𝑥2𝑏2 + 𝑦2𝑎2 = 𝑎2𝑏2 Ellipsning xossalari. yoki 1. Agarda M(x,y) nuqtaning 𝑥2 𝑦2 koordinatalari (2.4) tenglamani + = 1 𝑎2 𝑏2 qanoatlantirsa, u holda M (x,-y), M (-x,y), 1 2 (2.4) M (-x,-y), nuqtalarning koordinatalari ham 3 (2.4) tenglamani qanoatlantiradi. Bundan Xosil bo‘lgan (2.4) tenglama esa ellips ox o‘qiga, oy o‘qiga va O(0,0) ellipsning kanonik tenglamasi deyiladi. koordinata boshiga nisbat