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IKKI NOMA LUMLI TENGLAMANING GEOMETRIK
MANOSI
Sharipova Madina Po'latovna
Osiyo Xalqaro Universiteti "Umumtexnik fanlar" kafedrasi o'qituvchisi saripovamadina807.m@gmail.com
ARTICLE INFO
ABSTRACT
Qabul qilindi: 10-February 2024 yil Ma'qullandi: 15- February 2024 yil Nashr qilindi: 22- February 2024 yil
KEY WORDS
Chiziqli tenglamalar, to'gri chiziq ,parabola,aylana.
Maqolada ikki o'zgaruvchili tenglama va tenglamalar sistemasining geometrik o'rni ,ularning mohiyati o'rganilgan.
Ikki o'zgaruvchili tenglamalar sistemasi, matematikda, texnikada va boshqa sohalar bo'yicha juda muhim ahamiyatga ega.Ikki o'zgaruvchili tenglamalar sistemasi, har bir x qiymatiga mos keladigan y qiymatlarini hisoblash uchun amalga oshiriladi. Bu, grafiklar, ko'ordinatalar va boshqa usullar orqali ham yoritilishi mumkin. Umumiy qilib aytganda, har qanday ikki x va y noma'lum ga bog'liq bo'lgan tenglama tekislikda shunday nuqtalarning geometrik o'rnini bildiradiki, bu nuqtalarning koordinatalari shu tenglamani qanoatlantiradi. x va y ga bog'liq bo'lgan tenglamani
F(x,y)=0 (1)
ko'rinishida yozish mumkin. Bu tenglama qanday bo'lganda, qanday chiziq aniqlanishini misollarda ko'rib chiqamiz.
1. (1) tenglama Ax+By+C=0 ko'rinishidagi chiziqli tenglama bo'l-sin. Bu holda o'zgaruvchi x ning har bir qiymatiga y ning bitta qiymati mos keladi. Bunday (x,y) juftlardan bir nechtasini topib tekislikda belgi-laymiz va ularni tutashtirib, to'g'ri chiziqni hosil qilamiz.
Misol. 1) y=x tenglama birinchi va uchinchi koordinatalar burcha-gining bissektrisasini bildiradi (AB to'g'ri chiziq).
2. y=-x tenglama esa ikkinchi va to'rtinchi koordinatalar burchagi-ning bissektrisasini aniqlaydi (CD to'g'ri chiziq) (1-rasm).
1-rasm. 2-rasm.
Tenglamada o'zgaruvchilardan faqat bittasi qatnashishi mumkin. Bu holda ham tenglama biror chiziqni bildiradi.
Misol. x+2=0 tenglama berilgan bo'lsin.
Bundan x=-2 ni topamiz. Bu tenglama shunday nuqtalarning geometrik o'rnini aniqlaydiki, ularning har birining abssissasi x=-2 bo'lib, ordinatasi ixtiyoriy bo'ladi, bunday nuqtalar abssissa o'qidan - 2 ga teng nuqtadan o'tadi va 0y o'qiga parallel bo'lgan to'g'ri chiziq bo'ladi (2-rasm).
Shunga o'xshash, y - 3=0 tenglama ordinata o'qidan 3 ga teng kesmani ajratuvchi va 0x o'qiga parallel bo'lgan to'g'ri chiziqni bildiradi (3-rasm).
1 y y 3
0 x
OVA
1ЛЭЕ
-112
3-rasm. 4-rasm.
2. Ikkinchi darajali noma'lum qatnashgan tenglamani ko'rib chiqamiz. Misol. 1) x2-y=0 tenglama uchi koordinatalar boshida va tarmoqlari yuqoriga qaragan parabolani bildiradi (4-rasm).
2) x2+y2=4 tenglama markazi koordinatalar boshida, radiusi R=2 bo'lgan aylanani bildiradi (5-rasm).
x
5-rasm.
3. Agar (1) tenglamaning chap tomoni ko'paytuvchilarga ajralsa, har bir ko'paytuvchini alohida-alohida nolga tenglashtirib, bir nechta chiziqlarni hosil qilamiz.
Misol. x2-y2=0 yoki (x+y) (x-y)=0 tenglama x+y=0 va x-y=0 to'g'ri chiziqlar juftini aniqlaydi.
1. Xususiy holda F(x,y)=0 tenglama bitta yoki bir nechta nuqtalar-dan iborat bo'lgan to'plamni aniqlashi mumkin.
Misol. x2+y2=0 tenglama faqat 0(0,0) nuqtani ifodalaydi (x2-4)2+(y2-1)2=0 tenglama to'rtta nuqta (-2;-1), (-2;1), (2;-1), (2;1) ni aniq-laydi.
5. F(x,y)=0 tenglama bironta ham nuqtani aniqlamasligi mumkin. Misol, x2+y2+1=0 tenglamani haqiqiy sonlar juftining birontasi ham qanoat-lantirmaydi, demak bu tenglamaga hech qanday nuqta mos kelmaydi.
\ax+by+c = o
Chiziqli < tenglamalar sistemasi berilgan bo'lsin.
[a2 x + b2 y + c2 = 0
Ma'lumki, sistemadagi har bir tenglama to'g'ri chiziqni bildiradi. Sistemaning yechimi ikkala to'g'ri chiziqqa umumiy bo'lgan nuqtasi-ning koordinatalaridan iborat bo'ladi. Bu nuqta to'g'ri chiziqlarning ke-sishish nuqtasidir. Isbotsiz quyidagini keltiramiz.
a b
1. Agar — ^ — bo'lsa, sistema yagona yechimga ega boladi (to'g'ri chiziqlar
a2 b2
kesishadi).
2. Agar — = — ^ — bo'lsa, sistema yechimga ega emas (to'g'ri chi-ziqlar parallel)
a b c
'2 b2 C2
ll II lC
a2 b2 C2
chiziqlar ustma-ust tushadi).
Í2 x - y + 5 = 0 2 -1
Misol. 1) < sistema yagona yechimga ega, chunki —
[x + 3y - 7 = 0 13
|x - y + 3 = 0 1 -1 3
2) < sistema yechimga ega emas, chunki — = — ^ —
[2x - 2y - 5 = 0 2 - 2 - 5
|2x - 3 y +1 = 0 2 - 3 1
3) < sistema cheksiz ko'p yechimga ega, chunki — = — = —
[- 2x + 3 y -1 = 0 - 2 3 -1
Agar
biror chiziqni anglatadi. Sistemaning yechimi esa bu chi-ziqlarning kesishish nuqtalarining koordinatalaridan iborat bo'ladi.
Yechish: x+y=1 to'g'ri chiziq va x2+y2=4 aylanani bitta chizmada tasvirlaymiz. Ularning kesishish nuqtalari A va B ning koordinatalari sistemaning yechimi boladi. Demak, sistema 2 ta yechimga ega ekan (6-rasm).
u y
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