ikkinchi tartibli egri chiziqlar

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1576494378.doc 1 b y a x 2 2 2 2 = + a c = e 1 b ) y y ( a ) x x ( 2 2 0 2 2 0 = - + - x 8 y ) 2 x ( 2 2 2 - = + - 1 12 y 16 x 2 2 = + 1 2 2 2 2 = - b y a x x a b y ± = 1 a c > = e 1 b ) y y ( a ) x x ( 2 2 0 2 2 0 = - - - 2 x 2 1 y ± = 1 b y a x 2 2 2 2 = - 2 1 a b = ï î ï í ì = + = 25 b a 2 1 a b 2 2 5 2 a = 5 2 …

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yig`indisi o`zgarmas kattalikka (fokuslar orasidagi masofadan katta) teng nuqtalar to`plamiga ellips deyiladi. agar o`zgarmas kattalikni 2a, fokuslar orasidagi masofani esa 2c bilan belgilasak va tekislikda ox abssissa o`qi fokuslari orqali o`tuvchi, koordinatalar boshi f1f2 kesmaning o`rtasida joylashgan koordinatalar sistemasi tanlasak, ellips tenglamasi soddalashadi va quyidagi kanonik ko`rinishga keladi , bu yerda b 2 = a 2 – c 2 ( a > c ). ushbu holda ellips fokuslari: f1(-c; 0), f2(c; 0) (2-rasm ). koordinatalar boshi 0 nuqta ellipsning simmetriya markazi, koordinata o`qlari esa uning simmetriya o`qlari hisoblanadi. a1(-a; 0), a2(a; 0), b1(0; -b), b2(0; b) nuqtalarga ellipsning uchlari, 0a2 = a va 0a1= b kesma uzunliklariga uning mos ravishda katta va kichik yarim o`qlari deyiladi. shunday qilib, ellips ikki simmetriya o`qlariga va simmetriya markaziga ega qavariq yopiq chiziqdir. kattalikka ellipsning ekstsentrisiteti deb ataladi va har qanday ellips uchun ε a). giperbola fokuslari: f1(-c; 0) va f2(c; 0) (4-rasm). 0 …

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qlari koordinata o`qlariga parallel giperbola tenglama bilan aniqlanadi. yarim o`qlari teng, ya`ni a = b giperbolaga teng tomonli giperbola deyiladi. teng tomonlama giperbola tenglamasi x2 – y2 = a2 ko`rinishda bo`lib, uning asosiy to`g`ri to`rtburchagi kvadratdan iborat va ekstsentrisiteti ga teng. 4-rasm. 5-rasm. masala. asimptotalari tenglamalar bilan berilgan, fokus-lari orasidagi masofa 10 birlikka teng bo`lgan giperbola tenglamasini tu-zing. giperbola fokuslari abssissa o`qida yotadi deb qarab, uning kanonik tenglamasini tuzamiz: . fokuslar orasidagi masofa f1f2 = 2c = 10 bo`lganidan, c = 5. giperbola uchun c2 = a2 + b2 bo`lganidan va berilganidan foydalanib, quyidagi sistemani tuzamiz va uni yechamiz: sistema yechimi: va . demak, giperbola tenglamasi kanonik tenglamadan iborat (5–rasm). parabola va uning kanonik tenglamasi tekislikda fokusi deb ataluvchi berilgan f nuqtadan va direktrisasi deb ataluvchi berilgan dd( to`g`ri chiziqdan teng masofada yotuvchi nuqtalar tuplamiga parabola deyiladi. abssissa o`qi f fokus nuqtadan dd( direktrisaga perpendikulyar ravishda o`tuvchi, ordinata o`qi esa …

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. masala shartiga binoan km = am (7-rasm). berilgan aylana radiusi 0k=2 ekanligini va km = 0m – 0k tenglikni hisobga olsak, koordinatalarda quyidagi tenglamani olamiz: x2 + y2 - 2 = |x| yoki y 2 = 4 |x| + 4. ushbu tenglama uchlari (-1; 0) va (1; 0) nuqtalarda, fokuslari koordinatalar boshida, direktrisalari mos ravishda x = -2 va x = 2 to`g`ri chiziqlardan iborat, abssissa o`qi simmetriya o`qi bo`lgan parabolalarni ifodalaydi (7-rasm). y y b a 0 r x x m000 m x=8 4 m 0 2 x y d m a1 a2 b2 b1 f2 f1 y x 0 0 y x a1 a2 b2 0 x y f2 f1 b1 d d( 0 f x y a x k m y 0 _1406379574.unknown _1406379578.unknown _1406379582.unknown _1406379584.unknown _1406379586.unknown _1406379587.unknown _1406379585.unknown _1406379583.unknown _1406379580.unknown _1406379581.unknown _1406379579.unknown _1406379576.unknown _1406379577.unknown _1406379575.unknown _1406379570.unknown _1406379572.unknown _1406379573.unknown _1406379571.unknown _1406379568.unknown _1406379569.unknown _1406379567.unknown

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1576494378.doc 1 b y a x 2 2 2 2 = + a c = e 1 b ) y y ( a ) x x ( 2 2 0 2 2 0 = - + - x 8 y ) 2 x ( 2 2 2 - = + - 1 12 y 16 x 2 2 = + 1 2 2 2 2 = - b y a x x a b y ± = 1 a c > = e 1 b ) y y ( a ) x x ( 2 2 0 2 2 0 = - - - 2 x 2 1 y ± = 1 b y a x 2 2 2 2 …

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