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Anal ytická geometria kvadratických útvarov

Anal ytická geometria kvadratických útvarov. Romana Fabiniová. Kužeľosečky Analytické vyjadrenie kružnice a kruhu. k= { X є R 2 ; I XS I = r }. y. k : ( x – s 1 ) 2 + ( y – s 2 ) 2 = r 2. X. y. stredová rovnica kružnice. s 2. S. x. s 1. x.

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Anal ytická geometria kvadratických útvarov

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  1. Analytická geometria kvadratických útvarov Romana Fabiniová

  2. KužeľosečkyAnalytické vyjadrenie kružnice a kruhu k= { Xє R2;IXSI = r } y k : (x – s1)2+ (y – s2 ) 2 = r 2 X y stredová rovnica kružnice s2 S x s1 x

  3. Všeobecná rovnica kružnice : x2 + y2 + ax + by + c = 0 1.Príklad : urč S a r danej kružnice: k: x2+ y2 – 6x + 4y – 12 = 0 x2 –6x + y2 + 4y –12 = 0 x2 – 6x + 9 – 9 + y2 + 4y + 4 – 4 – 12 = 0 ( x – 3 )2 + ( y + 2 )2 –25 = 0 ( x – 3 )2 + ( y + 2 )2 = 25 => S [ 3 , - 2] r = 5 2. Príklad : k: x2 + y2 + 6x – 4y – 23 = 0

  4. Riešenie: x2 + y2 + 6x – 4y – 23 = 0 x2 + 6x + y2 – 4y – 23 = 0 x2 + 6x + 9 – 9 + y2 – 4y + 4 – 4 – 23 = 0 (x + 3)2 + (y – 2)2 – 36 = 0 (x + 3)2 + (y – 2 )2 = 36 S[ 3,2] r = 6

  5. Vzájomná poloha priamky a kružnice A * * S * S S T k B k k B s n t p ∩ k = 0 nesečnica p ∩ k = {T} dotyčnica p ∩ k = { A, B} sečnica p: ax + by + c = 0 k: x2 + y2 +ax + by + c = 0 - hľadáme spoločné body priamky a kružnice =>riešime sústavu lineárnej a kvadratickej rovnice ak D < 0 => p je nesečnica D > 0 => p je dotyčnica D = 0 => p je sečnica

  6. 1.Príklad: p : 2x – y = 0 => y = 2x k: x2 + y2 – 3x + 2y – 3 = 0 X2 + ( 2x)2 – 3x + 2 (2x) -3 = 0 X2 + 4x2 – 3x + 4y – 3 = 0 5x2 + x – 3 = 0 D = b2 – 4ac = 1 -4.5.(-3)= 1+60 = 61 x ½ = -1 +/- 7,8 = 0,68 10 -0,88 p je sečnica A : y = 2.0,68 = 1,36 A[ 0,68 ; 1,36 ] B : y = 2.(-0,88) = - 1,76 B[-0,88 ; - 1,76] 2.Príklad : p: 2x – y – 6 = 0 k: x2 + y2 – 4x – 5y – 1 = 0

  7. Riešenie : p: 2x – y – 6 = 0 =>y = 2x - 6 k: x2 + y2 – 4x – 5y – 1 = 0 k: x2 + (2x – 6)2 – 4x – 5(2x – 6) – 1 = 0 x2 + 4x2 – 24x + 36 – 4x – 10x + 30 – 1 = 0 5x2 – 38x + 65 = 0 D = b2 – 4.a.c = 1444 – 4.5.65 = 144 x1/2 = 2,6 5 y1 = - 0,8 y2 = 4 A [ 2,6 ; - 0,8] B [ 5 , 4 ]

  8. Vzájomná poloha 2 kružníc 2. 3. S1 A 1. T I S1 S2 I * * r1 S2 S1 S2 r2 r2 * * r1 S2 S1 B r1 + r2 I S1 S2 I r1 + r2 = I S1 S2 I r1 + r2 I S1 S2 I k1∩ k2 = { A , B} * T * S2 S1 * * * S2 * S1 * S1 = S2 k1 ∩ k2 = 0 k1∩k2 = {T} k1 = k2

  9. Príklad : Vyšetri vzájomnú polohu týchto útvarov : k1 : x2 + y2 - 4x -18y + 75 = 0 } - k2 : x2 + y2 – 20y + 75 = 0 / .(-1) Úloha : nájdite spoločné body - 4x + 2y = 0 / : 2 - 2x + y = 0 => k2: x2 + 4x2 – 40x + 75 = 0 5x2 – 40x + 75 = 0 / : 5 x2 – 8x + 15 = 0 ( x – 3).(x – 5 ) = 0 x1 = 3 x2 = 5 y = 2x y1 = 2.3 = 6 y2 = 2.5 = 10 A[ 3,6] B[ 5,10]

  10. Rovnica dotyčnice k : x2 + y2 + ax + by + c = 0...... S [ s1,s2] r = T [ xT, yT] u (S) = T u = nT p: ax + by + c = 0 Tє k * u S T t 1.Príklad: t = ? k: x2 + y2 = 25 T [ 4, -3] t : 4x – 3y + c = 0 Tє t: 16 + 9 + c = 0 c = -25 t : 4x – 3y – 25 = 0 T є k ? 16+9 = 25 → S [ 0,0] r = 5 u (S)= T u (4,-3)=nT → → 2.Príklad: k: x2 + y2 – 6x – 10y + 9 = 0 C[ - 1,2] Úloha: napíš rovnicu dotyčnice

  11. Riešenie: k: x2 + y2 – 6x – 10y + 9 = 0 C[ -1,2] u (S) = C u ( - 4, -3)= nt -4x – 3y + c = 0 Cєk: 4 – 6 + c = 0 c = 2 → k: x2 – 6x + y2 – 10y + 9 = 0 x2 – 6x + 9 – 9 + y2 – 10y + 25 – 25 + 9 = 0 (x – 3)2 + (y – 5)2 = 25 S[ 3 , 5] r = 5 → → t : - 4x – 3y + 2 = 0

  12. Rovnica dotyčnice ku kružnici z bodu Postup: 1. SI;SI SR ° T1 TK 2. TK; TK (SI, ISSII) 3.T1,T2; k ∩ TK = { T1,T2} 4.t1, t2; t1 = RT1 t2 = RT2 r t2 α 2 (α * * R S t1 SI k T2

  13. Guľová plocha , guľa Guľová plocha H = { X є R3 I XSI = r } X Guľa G = { X є R3 IXSI < r} S G H stredova rovnica : (x- s1)2 + (y – s2)2 + (z – s3)2 = r2 všeobecná rovnica: x2 + y2 +z2 + Ax +By + Cz + D = 0

  14. Elipsa C x E = { X є R2 ; IXF1I + IXF2I = 2a} S…..........stred elipsy F1,F2........ohniská A, B.........hlavné body C, D.........vedľajšie body a.............hlavná poloos a = IASI = IBSI b.............vedľajšia poloos b = ICSI = IDSI e.............ohnisková vzdialenosť= excentricita e = IF1SI = IF2SI b a A B F1 a e F2 S D b2 + e2 = a2

  15. a є oX a є oY ↑ ↑ x2 b2 y2 a2 x2 y2 a2 b2 y + = 1 y = 1 + stredová rovnica elipsy : S[ 0,0] → → S x S x S[ s1,s2] a є oX a є oY ↑ (x – s1)2 b2 (y – s2)2 a2 (x – s1)2 a2 (y – s2)2 b2 y = 1 ↑ + = 1 + y s2 S s2 ← S ↓ → → s1 x s1 x

  16. všeobecná rovnica elipsy A = 0 A,B > 0 Ax2 + By2 + Cx + Dy + E = 0 1.Príklad : -charakterizuj elipsu,ktorá má takúto rovnicu: 9x2 + 25y2 – 54x + 100y – 44 = 0 9x2 – 54x + 25y2 + 100y – 44 = 0 9(x2 – 6x + 9 – 9) + 25 (y2 + 4y + 4 – 4 ) – 44 = 0 9(x – 3)2 + 25(y + 2)2 – 100 – 44 = 0 9(x – 3)2 + 25(y + 2)2 = 225 /:225 9(x – 3)2 225 2.príklad: 36x2 + 100y2 + 72x – 600y – 2664 = 0 a = 5 b = 3 e = 4 25(y + 2)2 225 + = 1 (x – 3)2 25 (y + 2)2 9 = 1 =>a // osX +

  17. Riešenie : 36x2 + 100y2 + 72x – 600y – 2664 = 0 36x2 + 72x + 100y2 – 600y – 2664 = 0 36(x2 + 2x) + 100(y2 – 6y) – 2664 = 0 36(x2 + 2x + 1 – 1) + 100(y2 - 6y + 9 – 9) – 2664 = 0 36(x + 1)2 – 36 + 100(y – 3)2 -900 – 2664 = 0 36(x + 1)2 + 100(y – 3)2 = 3600 36(x + 1)2 3600 100(y – 3)2 3600 = 1 + (x + 1)2 100 (y – 3)2 36 = 1 • a // os X + a = 10 b = 6

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