Reja: Markazli egri chiziqning tenglamasini soddalashtirish

Umumiy tenglama bilan berilgan ikkinchi tartibli chiziqlarni aniqlash va sinflarga ajratish.

𝑥^′ = 𝑥 + 𝑎
^{𝑦^′ = 𝑦 + 𝑎
parallel ko‘chirish formulasi.
𝑥 = 𝑥^′𝑐𝑜𝑠𝛼 − 𝑦^′𝑠𝑖𝑛𝛼

burish formulasi.

{
𝑦 = 𝑥
^′𝑠𝑖𝑛𝛼 + 𝑦
^′𝑐𝑜𝑠𝛼

𝑥 = 𝑥^′𝑐𝑜𝑠𝛼 − 𝑦^′𝑠𝑖𝑛𝛼 + 𝑎
^{𝑦 = 𝑦^′𝑐𝑜𝑠𝛼 + 𝑥^′𝑠𝑖𝑛𝛼 + 𝑏
parallel ko‘chirish va burish birgalikda harakat deyiladi.
𝑎₁₁𝑥² + 𝑎₂₂𝑦² + 2𝑎₁₀𝑥 + 2𝑎₂₀𝑦 + 2𝑎₁₂𝑥𝑦 + 𝑎₀₀ = 0
shu ifoda bilan berilgan tenglama ikkinchi tartibli chiziqning umumiy
tenglamasi deyiladi.
Umumiy tenglama bilan berilgan ikkinchi tartibli chiziqni qanday chiziq ekanligini aniqlash uchun quyidagi ishlarni bajaramiz.
𝑥^′ = 𝑥 + 𝑎
^{𝑦^′ = 𝑦 + 𝑎
𝑎₁₁(𝑥^′ + 𝑎)² + 𝑎₂₂(𝑦^′ + 𝑏)² + 2𝑎₁₀⁽𝑥^′ + 𝑎⁾ + 2𝑎₂₀⁽𝑦^′ + 𝑏⁾ +

22

10
+2𝑎₁₂⁽𝑥^′ + 𝑎⁾⁽𝑦^′ + 𝑏⁾ + 𝑎₀₀ = 0 ⟹

^𝑎11
(𝑥^′2 + 2𝑎𝑥^′ + 𝑎²) + 𝑎
(𝑦^′2 + 2𝑏𝑦^′ + 𝑏²) + (2𝑎
𝑥^′ + 2𝑎𝑎₁₀) +

+2𝑎₂₀𝑦^′ + 2𝑎₂₀𝑏 + 2𝑎₁₂⁽𝑥^′𝑦^′ + 𝑎𝑦^′ + 𝑏𝑥^′ + 𝑎𝑏⁾ + 𝑎₀₀ = 0 ⟹
𝑎₁₁𝑥^′2 + 2𝑎𝑎₁₁𝑥^′ + 𝑎²𝑎₁₁ + 𝑎₂₂𝑦^′2 + 2𝑦^′𝑏𝑎₂₂ + 𝑎₂₂𝑏² +
+2𝑎₁₀𝑥^′ + 2𝑎𝑎₁₀ + 2𝑎₂₀𝑦^′ + 2𝑎₂₀𝑏 + 2𝑎₁₂𝑥^′𝑦^′ +
+2𝑎₁₂𝑎𝑦^′ + 2𝑎₁₂𝑏𝑥^′ + 2𝑎₁₂𝑎𝑏 + 𝑎₀₀ = 0

_{2𝑎₁₁𝑎 + 2𝑎₁₂𝑏 = −2𝑎₁₀ 2𝑎₂₂𝑏 + 2𝑎₁₂𝑎 = −2𝑎₂₀

𝑎₁₁𝑥^′2
+ 𝑎₂₂𝑦^′2
𝑥 = 𝑥^′′𝑐𝑜𝑠𝛼 − 𝑦^′′𝑠𝑖𝑛𝛼
^{+ 𝐴00 = 0 ⟹ {}𝑦 = 𝑥^′′𝑠𝑖𝑛𝛼 + 𝑦^′′𝑐𝑜𝑠𝛼

kelib chiqadi.
𝑎₁₁(𝑥^′′𝑐𝑜𝑠𝛼 − 𝑦^′′𝑠𝑖𝑛𝛼)² + 𝑎₂₂(𝑥^′′𝑠𝑖𝑛𝛼 + 𝑦^′′𝑐𝑜𝑠𝛼)² +
+2𝑎₁₂⁽𝑥^′′𝑐𝑜𝑠𝛼 − 𝑦^′′𝑠𝑖𝑛𝛼⁾⁽𝑥^′′𝑠𝑖𝑛𝛼 + 𝑦^′′𝑐𝑜𝑠𝛼⁾ + 𝐴₀₀ = 0
𝑎₁₁(𝑥^′′2𝑐𝑜𝑠²𝛼 − 2𝑎₁₁𝑥^′′𝑐𝑜𝑠𝛼𝑦^′′𝑠𝑖𝑛𝛼 + 𝑦^′′2𝑠𝑖𝑛²𝛼) +
+𝑎₂₂(𝑥^′′2𝑠𝑖𝑛²𝛼 + 2𝑥^′′𝑠𝑖𝑛𝛼𝑦^′′𝑐𝑜𝑠𝛼 + 𝑦^′′2𝑐𝑜𝑠²𝛼) + 𝐴₀₀ = 0 ⟹
𝑎₁₁𝑥^′′2𝑐𝑜𝑠²𝛼 − 2𝑎₁₁𝑥^′′𝑐𝑜𝑠𝛼𝑦^′′𝑠𝑖𝑛𝛼 + 𝑎₁₁𝑦^′′2𝑠𝑖𝑛²𝛼 +
+𝑎₂₂𝑦^′′2𝑐𝑜𝑠²𝛼 + 2𝑎₁₃𝑥^′′2𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼 + 2𝑎₁₂𝑥^′′2𝑦^′′2𝑐𝑜𝑠²𝛼 −
−2𝑎₁₂𝑥^′′𝑦^′′𝑠𝑖𝑛²𝛼 − 2𝑎₁₂𝑦^′′2𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼 + 𝐴₀₀ = 0 ⟹
2
(𝑎₁₁𝑐𝑜𝑠²𝛼 + 𝑎₂₂𝑠𝑖𝑛²𝛼 + 2𝑎₁₂𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼)𝑥^′′ +
2
+(𝑎₁₁𝑠𝑖𝑛²𝛼 + 𝑎₂₂𝑐𝑜𝑠²𝛼 − 2𝑎₁₂𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼)𝑦^′′ +

+⁽−2𝑎₁₁𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼 + 2𝑎₂₂𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼 + 2𝑎₁₂𝑐𝑜𝑠²𝛼
− 2𝑎₁₂𝑠𝑖𝑛²𝛼⁾𝑥^′′𝑦^′′ + 𝐴₀₀ = 0 ⁽𝑎₂₂ − 𝑎₁₁⁾𝑠𝑖𝑛2𝛼 + 2𝑎₁₂𝑐𝑜𝑠2𝛼 = 0 ⟹
𝑡𝑔2𝛼 = ^2𝑎12
𝑎11 ^{− 𝑎}22

(5.42)

𝐴₁₁𝑥^′2 + 𝐴₂₂𝑦^′2 + 𝐴₀₀ = 0 tenglamani xususiy hollarini qaraymiz:
1) 𝐴₁₁ > 0, 𝐴₂₂ > 0, 𝐴₀₀ < 0 bo‘lsa, ellips;
2) 𝐴₁₁ < 0, 𝐴₂₂ < 0, 𝐴₀₀ > 0 bo‘lsa, ellips;
3) 𝐴₁₁ 𝐴₂₂ < 0, 𝐴₀₀ > 0 bo‘lsa, giperbola;
4) 𝐴₁₁ 𝐴₂₂ < 0, 𝐴₀₀ < 0 bo‘lsa, giperbola;
5) 𝐴₁₁ 𝐴₂₂ < 0, 𝐴₀₀ = 0 bo‘lsa, nuqta;
6) 𝐴₁₁ = 0, 𝐴₂₂ 𝐴₀₀ < 0, 𝐴₀₀ > 0 bo‘lsa, parallel to‘g‘ri chiziqlar hosil bo‘ladi.
1-Misоl. Quyidagi tenglamaning geometrik ma’nosini tekshirib, uning kanonik tenglamasi tuzilsin:
5𝑥² + 4𝑥𝑦 + 8𝑦² − 32𝑥 − 56𝑦 + 80 = 0.
Yechish: Egri chiziqning jinsini aniqlash uchun ∆ va 𝑀 ni hisoblashga to‘g‘ri keladi:
5 2 − 16

∆= |
_{2 8 − 28} | = −1296, 𝐴 ∙ ∆< 0.
−16 − 28 80

𝑀 = 𝐵² − 𝐴𝐶 = 4 − 5 ∙ 8 = −36 < 0.
Demak, berilgan tenglama haqiqiy ellipsdan iborat. Uning kanonik tenglamasining ko‘rinishi:

𝐴₁
𝑥² + 𝐶₁
𝑦² = ^∆ ,
𝑀

𝐴₁
1
=
2
= ¹ [𝐴 + 𝐶 + √⁽𝐴 − 𝐶⁾² + 4𝐵²]=
2

[5 + 8 + ^√(5 − 8⁾² + 4 ∙ 2²] = 9;

1

𝐶₁ =
1
[𝐴 + 𝐶 − ^√(𝐴 − 𝐶⁾² + 4𝐵²] =
2

= [5 + 8 − ^√(5 − 8⁾² + 4 ∙ 2²] = 4;
2

∆ −1296
=
𝑀 −36
= 36.

Demak, ellipsning kanonik tenglamasi
9𝑥² + 4𝑦² = 36

yoki


_𝑥2



4


_𝑦2
+
9

= 1.

2-Misоl. 𝑂𝑥 o‘qi parabolaning simmetriya o‘qi bo‘lib, uning uchi
koordinatalar boshida yotadi. Parabola uchidan fokusigacha bo‘lgan masofa 4 birlikka tеng. Parabola va uning direktrisasi tenglamasini toping.
Yechish: Dastlab, masala shartiga asosan, parabolaning 𝑝
parametrini topamiz:
^|𝑂𝐹^| = 4 ⟹ 𝑝/2 = 4 ⟹ 𝑝 = 8.
Unda, (5.39) formulaga asosan, parabola tenglamasini topamiz:
𝑦² = 2𝑝𝑥 ⟹ 𝑦² = 2 ∙ 8𝑥 = 16𝑥.
Bu yerdan direktrisa tenglamasi 𝑥 = −𝑝/2 ⟹ 𝑥 = −4 ekanligini ko‘ramiz.
Shuni ta’kidlab o‘tish kerakki, 𝑦 = 𝑎𝑥² + 𝑏𝑥 + 𝑐 (𝑎 ≠ 0) kvadrat uchhadning grafigi uchi koordinatalari

x₀   ^b ,
2a

24. 
    _ 4ac  b^{2 0} 4a
    

bo‘lgan 𝑀₀(𝑥₀; 𝑦₀) nuqtada, simmetriya o‘qi esa 𝑂𝑦 o‘qiga parallel va
𝑥 = −𝑏/2𝑎 tenglamaga ega bo‘lgan vertikal to‘g‘ri chiziqdan tashkil

topgan paraboladan iboratdir. Agar 𝑎 > 0 bo‘lsa, parabola yuqoriga,
𝑎 < 0 bo‘lsa, pastga yo‘nalgan bo‘ladi.

Mustaqil yechish uchun topshiriqlar.

1. 
   Quyidаgi gipеrbоlаlаrning tеnglаmаlаri sоddа shаklgа kеltirilsin:
   

1) 9𝑥² − 25𝑦² − 18𝑥 − 100𝑦 − 316 = 0;
2) 5𝑥² − 6𝑦² + 10𝑥 − 12𝑦 − 31 = 0;
3) 𝑥² − 4𝑦² + 6𝑥 + 5 = 0.
Mаrkаzlаrining kооrdinаtаlаri vа o‘qlаri tоpilsin.

2. 
   Quyidagi tenglamalar bilan qanday egri chiziqlar berilganligi tekshirilsin:
   

1) 𝑥² − 2𝑥𝑦 + 2𝑦² − 4𝑥 − 6𝑦 + 3 = 0;
2) 𝑥² − 2𝑥𝑦 − 2𝑦² − 4𝑥 − 6𝑦 + 3 = 0;
3) 𝑥² − 2𝑥𝑦 + 𝑦² − 4𝑥 − 6𝑦 + 3 = 0;
4) 𝑥² − 2𝑥𝑦 + 2𝑦² − 4𝑥 − 6𝑦 + 29 = 0;
5) 𝑥² − 2𝑥𝑦 − 2𝑦² − 4𝑥 − 6𝑦 − ¹³ = 0.
3

3. 
   Quyidagi egri chiziqlarning turlari aniqlansin:
   

1) 𝑥² + 6𝑥𝑦 + 𝑦² + 6𝑥 + 2𝑦 − 1 = 0;
2) 3𝑥² − 2𝑥𝑦 + 3𝑦² + 4𝑥 + 4𝑦 − 4 = 0;
3) 𝑥² − 4𝑥𝑦 + 3𝑦² + 2𝑥 − 2𝑦 = 0;
4) 𝑦² + 5𝑥𝑦 − 14𝑥² = 0;
5) 𝑥² − 𝑥𝑦 − 𝑦² − 𝑥 − 𝑦 = 0.

4. 
   Berilgan tenglamalarning chap tomonlarini ko‘paytuvchilarga ajratishdan foydalanib, tenglamalarning geometrik ma’nosi ko‘rsatilsin:
   

1) 𝑥𝑦 − 𝑏𝑥 − 𝑎𝑦 + 𝑎𝑏 = 0;
2) 𝑥² − 2𝑥𝑦 + 5𝑥 = 0;
3) 𝑥² − 4𝑥𝑦 + 4𝑦² = 0;
4) 9𝑥² + 30𝑥𝑦 + 25𝑦² = 0;
5) 4𝑥² − 12𝑥𝑦 + 9𝑦² − 25 = 0.
5.1.5. 𝑦² − 𝑥𝑦 − 5𝑥 + 7𝑦 + 10 = 0 tenglamaning ikki (qo‘sh) to‘g‘ri chiziqni ifodalashi tekshirilsin va bu to‘g‘ri chiziqlardan har birining tenglamasi topilsin.

6. 
   Quyidagi tenglama bilan berilgan ikkita to‘g‘ri chiziqdan har birining tenglamasi topilsin:
   

1) 21𝑥² + 𝑥𝑦 − 10𝑦² = 0;
2) 𝑥² + 2𝑥𝑦 + 𝑦² + 2𝑥 + 2𝑦 − 4 = 0;
3) 𝑦² − 4𝑥𝑦 − 5𝑦² + 5𝑥 − 𝑦 = 0;
4) 4𝑥² − 4𝑥𝑦 + 𝑦² + 12𝑥 − 6𝑦 + 9 = 0.

6. 
   Egri chiziqlar tekshirilsin:
   

1) 2𝑥² + 3𝑥𝑦 − 5𝑦² = 0;
2) 𝑥² + 4𝑥𝑦 + 4𝑦² = 0;
3) 10𝑥² − 7𝑥𝑦 + 𝑦² = 0;
4) 5𝑥² − 4𝑥𝑦 + 𝑦² = 0.

6. 
   Invariantlardan foydalanib, quyidagi egri chiziq tenglamalari sodda shaklga keltirilsin:
   

1) 𝑥² + 2𝑥𝑦 − 𝑦² + 8𝑥 + 4𝑦 − 8 = 0;
2) 7𝑥² − 24𝑥𝑦 − 38𝑥 + 24𝑦 + 175 = 0;
3) 5𝑥² + 8𝑥𝑦 + 5𝑦² − 18𝑥 − 18𝑦 + 9 = 0;
4) 5𝑥² + 12𝑥𝑦 − 22𝑥 − 12𝑦 − 19 = 0;
5) 6𝑥𝑦 + 8𝑦² − 12𝑥 − 26𝑦 + 11 = 0.
Bu egri chiziqlarning hammasi to‘g‘ri burchakli koordinatalar sistemasiga nisbatan berilgan.

6. 
   Invariantlardan foydalanib, quyidagi parabolalarning tenglamalari soddalashtirilsin:
   

1) 𝑥² − 2𝑥𝑦 + 𝑦² − 10𝑥 − 6𝑦 + 25 = 0;
2) 4𝑥² − 4𝑥𝑦 + 𝑦² − 2𝑥 − 14𝑦 + 7 = 0;
3) 𝑥² − 2𝑥𝑦 + 𝑦² − 𝑥 − 2𝑦 + 3 = 0;
4) 4𝑥² − 4𝑥𝑦 + 𝑦² − 𝑥 − 2 = 0.

𝜔 = ^𝜋
2
bo‘lgan hol uchun.

6. 
   Quyidagi egri chiziqlarning tenglamalari soddalashtirilsin:
   

1) 𝑥² − 3𝑥𝑦 + 𝑦² + 1 = 0, 𝜔 = 60⁰;
2) 2𝑥² + 2𝑦² − 2𝑥 − 6𝑦 + 1 = 0, 𝜔 = 60⁰;
3) 4𝑥² − 4𝑥𝑦 + 𝑦² − 4𝑥 − 4𝑦 + 7 = 0, 𝜔 = 120⁰.
5.1.11. Invariantlardan foydalanib, 8𝑦² + 6𝑥𝑦 − 12𝑥 − 26𝑦 + 11 = 0
giperbolaning asimptotalariga nisbatan tenglamasi yozilsin. 𝜔 = 90⁰.

12. 
    To‘g‘ri burchakli koordinatalar sistemasiga nisbatan quyidagi tenglamalar bilan berilgan giperbolalarning asimptotalariga nisbatan yozilgan tenglamasi topilsin:
    

1) 2𝑥² + 3𝑥𝑦 − 2𝑦² − 8𝑥 − 11𝑦 = 0;
2) 4𝑥² + 2𝑥𝑦 − 𝑦² + 6𝑥 + 2𝑦 + 3 = 0;

3) 𝑦² − 2𝑥𝑦 − 4𝑥 − 2𝑦 − 2 = 0.

12. 
    Biror to‘g‘ri burchakli koordinatalar sistemasiga nisbatan egri chiziq 5𝑥² + 2𝑥𝑦 − 22𝑥 − 12𝑦 − 19 = 0 tenglama bilan ifodalanadi. Bu egri chiziqning o‘z uchiga nisbatan tenglamasi topilsin.
    
13. 
    Quyidagi tenglamalarning har biri ellipsni ifodalasa, uning markazi bo‘lgan С nuqtaning koordinatasi, yarim o‘qi, ekssentrisiteti va direktrisasi tenglamalarini tuzing:
    

1) 5𝑥² + 9𝑦² − 30𝑥 + 18𝑦 + 9 = 0;
2) 16𝑥² + 25𝑦² + 32𝑥 − 100𝑦 − 284 = 0;
3) 4𝑥² + 3𝑦² − 8𝑥 + 12𝑦 − 32 = 0.

12. 
    Quyidagi tenglamalar giperbola hosil qilishini tekshirib va uning markazi bo‘lgan C nuqtaning koordinatasini, yarim o‘qlarini, ekssentrisitetini, asimptota va direktrisa tenglamalarini tuzing:
    

1) 16𝑥² − 9𝑦² − 64𝑥 − 54𝑦 − 161 = 0;
2) 9𝑥² − 16𝑦² + 90𝑥 + 32𝑦 − 367 = 0;
3) 16𝑥² − 9𝑦² − 64𝑥 − 18𝑦 + 199 = 0.

12. 
    Quyidagi chiziqlardan qaysilari: 1) yagona markazga; 2) ko‘p markazlarga; 3) markazga ega emasligini aniqlang.
    

1) 3𝑥² − 4𝑥𝑦 − 2𝑦² + 3𝑥 − 12𝑦 − 7 = 0;
2) 4𝑥² + 5𝑥𝑦 + 3𝑦² − 𝑥 + 9𝑦 − 12 = 0;
3) 4𝑥² − 4𝑥𝑦 + 𝑦² − 6𝑥 + 8𝑦 + 13 = 0;
4) 4𝑥² − 4𝑥𝑦 + 𝑦² − 12𝑥 + 6𝑦 − 11 = 0;
5) 𝑥² − 2𝑥𝑦 + 4𝑦² + 5𝑥 − 7𝑦 + 12 = 0;
6) 𝑥² − 2𝑥𝑦 + 𝑦² − 6𝑥 + 6𝑦 − 3 = 0;
7) 𝑥² − 20𝑥𝑦 + 25𝑦² − 14𝑥 + 2𝑦 − 15 = 0;
8) 4𝑥² − 6𝑥𝑦 − 9𝑦² + 3𝑥 − 7𝑦 + 12 = 0.

12. 
    Quyidagi berilgan chiziqlar markazga ega bo‘lsa, ularning markaziy nuqtalarini toping:
    

1) 3𝑥² + 5𝑥𝑦 + 𝑦² − 8𝑥 − 11𝑦 − 7 = 0;
2) 5𝑥² + 4𝑥𝑦 + 2𝑦² + 20𝑥 + 20𝑦 − 18 = 0;
3) 9𝑥² − 4𝑥𝑦 − 7𝑦² − 12 = 0;
4) 2𝑥² − 6𝑥𝑦 + 5𝑦² + 22𝑥 − 36𝑦 + 11 = 0.

12. 
    Quyidagi har bir chiziqning ko‘p markazli bo‘lishini tekshirib, ularning har biri uchun geometrik markazini aniqlaydigan tenglamasini tuzing:
    

1) 𝑥² − 6𝑥𝑦 + 9𝑦² − 12𝑥 + 36𝑦 + 20 = 0;
2) 4𝑥² + 4𝑥𝑦 + 𝑦² − 8𝑥 − 4𝑦 − 21 = 0;
3) 25𝑥² − 10𝑥𝑦 + 𝑦² + 40𝑥 − 8𝑦 + 7 = 0.

12. 
    Quyidagi tenglamalar markaziy chiziqni ifodalashini tekshirib, ularning har birini koordinata boshiga ko‘chiruvchi tenglamasini tuzing:
    

1) 3𝑥² − 6𝑥𝑦 + 2𝑦² − 4𝑥 + 2𝑦 + 1 = 0;
2) 6𝑥² + 4𝑥𝑦 + 𝑦² + 4𝑥 − 2𝑦 + 2 = 0;
3) 4𝑥² + 6𝑥𝑦 + 𝑦² − 10𝑥 − 10 = 0;
4) 4𝑥² + 2𝑥𝑦 + 6𝑦² + 6𝑥 − 10𝑦 + 9 = 0.

12. 
    т va п ning qanday qiymatlarida
    

𝑚𝑥² + 12𝑥𝑦 + 9𝑦² + 4𝑥 + 𝑛𝑦 − 13 = 0
tenglama quyidagilarni aniqlaydi:
а) markaziy chiziqni; б) markazsiz chiziqni;
в) ko‘p markazli chiziqlarni.

12. 
    Parallel ko‘chirish yo‘li bilan quyidagi tenglamalarning har birining turini aniqlab, sodda holga keltiring. Qanday geometrik shaklni ifodalashini toping. Eski va yangi koordinatalar sistemasida grafigini chizing.
    

1) 4𝑥² + 9𝑦² − 40𝑥 + 36𝑦 + 100 = 0;
2) 9𝑥² − 16𝑦² − 54𝑥 − 64𝑦 − 127 = 0;
3) 9𝑥² + 4𝑦² + 18𝑥 − 8𝑦 + 49 = 0;

12. 
    Quyidagi berilgan tenglamalarning har birini eng oddiy shaklga keltirib, ularning turini, qanday geometrik shaklni tasvirlashini, grafiklarning eski va yangi koordinata o‘qlariga nisbatan joylashishini aniqlang:
    

1) 32𝑥² + 52𝑥𝑦 − 7𝑦² + 180 = 0;
2) 5𝑥² − 6𝑥𝑦 + 5𝑦² − 32 = 0;
3) 17𝑥² − 12𝑥𝑦 + 8𝑦² = 0;
4) 5𝑥² + 24𝑥𝑦 − 5𝑦² = 0;
5) 5𝑥² − 6𝑥𝑦 + 5𝑦² + 8 = 0.

12. 
    Quyidagi berilgan tenglamalarning har birini eng oddiy shaklga keltirib, ularning turini, qanday geometrik shaklni tasvirlashini, grafiklarning eski va yangi koordinata o‘qlariga nisbatan joylashishini aniqlang:
    

1) 14𝑥² + 24𝑥𝑦 + 21𝑦² − 4𝑥 + 18𝑦 − 139 = 0;
2) 11𝑥² − 20𝑥𝑦 − 4𝑦² − 20𝑥 − 8𝑦 + 1 = 0;
3) 7𝑥² + 60𝑥𝑦 + 32𝑦² − 14𝑥 − 60𝑦 + 7 = 0;
4) 50𝑥² − 8𝑥𝑦 + 35𝑦² + 100𝑥 − 8𝑦 + 67 = 0;

12. 
    Quyidagi tenglamalarni koordinatalar sistemasini almashtirmasdan, har biri ellipsni ifodalashini va uning yarim o‘qlardagi qiymatlarini toping:
    

1) 41𝑥² + 24𝑥𝑦 + 9𝑦² + 24𝑥 + 18𝑦 − 36 = 0;
2) 8𝑥² + 4𝑥𝑦 + 5𝑦² + 16𝑥 + 4𝑦 − 28 + 9 = 0;
3) 13𝑥² + 18𝑥𝑦 + 37𝑦² − 26𝑥 − 18𝑦 + 3 = 0;
4) 13𝑥² + 10𝑥𝑦 + 13𝑦² + 46𝑥 + 62𝑦 + 13 = 0.

12. 
    Koordinatalar sistemasini almashtirmasdan quyidagi tenglamalar bitta nuqtani ifodalashini isbotlang.
    

1) 5𝑥² − 6𝑥𝑦 + 2𝑦² − 2𝑥 + 2 = 0;
2) 𝑥² + 2𝑥𝑦 + 2𝑦² + 6𝑦 + 9 = 0;
3) 5𝑥² + 4𝑥𝑦 + 𝑦² − 6𝑥 − 2𝑦 + 2 = 0;
4) 𝑥² − 6𝑥𝑦 + 10𝑦² + 10𝑥 − 32𝑦 + 26 = 0.

12. 
    Koordinatalar sistemasini almashtirmasdan, har biri giperbolani ifodalasa, uning yarim o‘qlarining qiymatini toping:
    

1) 4𝑥² + 24𝑥𝑦 + 11𝑦² + 64𝑥 + 42𝑦 + 51 = 0;
2) 12𝑥² + 26𝑥𝑦 + 12𝑦² − 52𝑥 − 48𝑦 + 73 = 0;
3) 3𝑥² + 4𝑥𝑦 − 12𝑥 + 16 = 0;
4) 𝑥² − 6𝑥𝑦 − 7𝑦² + 10𝑥 − 30𝑦 + 23 = 0.

12. 
    Koordinatalar sistemasini almashtirmasdan, quyidagi tenglamalarning har biri kesishgan to‘g‘ri chiziqlar juftligini (degenerat giperbola) belgilashini aniqlang va ularning tenglamalarini toping:
    

1) 3𝑥² + 4𝑥𝑦 + 𝑦² − 2𝑥 − 1 = 0;
2) 3𝑥² − 6𝑥𝑦 + 8𝑦² − 4𝑦 − 4 = 0;
3) 𝑥² − 4𝑥𝑦 + 3𝑦² = 0;
4) 𝑥² + 4𝑥𝑦 + 3𝑦² − 6𝑥 − 12𝑦 + 9 = 0.

12. 
    Quyidagi tenglamalarning har biri parabolik ekanligini aniqlang; ularning har birini eng oddiy shaklga keltiring; ular qanday geometrik tasvirlarni belgilashlarini aniqlang:
    

1) 9𝑥² + 24𝑥𝑦 + 16𝑦² − 18𝑥 + 226𝑦 + 209 = 0;
2) 𝑥² − 2𝑥𝑦 + 𝑦² − 12𝑥 + 12𝑦 − 14 = 0;
3) 4𝑥² + 12𝑥𝑦 + 9𝑦² − 4𝑥 − 6𝑦 + 1 = 0.

12. 
    Koordinatalar sistemasini almashtirmasdan, quyidagi tenglamalarning har biri parabolani aniqlaganligini aniqlang va ushbu parabola parametrini toping:
    

1) 9𝑥² + 24𝑥𝑦 + 16𝑦² − 120𝑥 + 90𝑦 = 0;
2) 9𝑥² − 24𝑥𝑦 + 16𝑦² − 54𝑥 − 178𝑦 + 181 = 0;
3) 𝑥² − 2𝑥𝑦 + 𝑦² + 6𝑥 − 14𝑦 + 29 = 0;
4) 9𝑥² − 6𝑥𝑦 + 𝑦² − 50𝑥 + 50𝑦 − 275 = 0.

12. 
    Koordinatalar sistemasini almashtirmasdan, quyidagi tenglamalarning har biri bir juft parallel to‘g‘ri chiziqni belgilashini aniqlang va ularning tenglamalarini toping:
    

1) 4𝑥² + 4𝑥𝑦 + 𝑦² − 12𝑥 − 6𝑦 + 5 = 0;
2) 4𝑥² − 12𝑥𝑦 + 9𝑦² + 20𝑥 − 30𝑦 − 11 = 0;
3) 25𝑥² − 10𝑥𝑦 + 𝑦² + 10𝑥 − 2𝑦 − 15 = 0.