Tiêu đề Conics Engg Question Bank Solution (HSC 26).pdf ✅

2  Higher Math 2nd Paper Chapter-6 5. (3, – 1) Ges (1, – 1) Dc‡K›`a wewkó Dce„‡Ëi GKwU kxl©we›`y n‡Z Dc‡K›`a؇qi `~i‡Z¡i ̧Ydj 4 GKK n‡j Dce„‡Ëi mgxKiY wbY©q Ki| [BUET 21-22] mgvavb: Dce„ËwUi †K‡›`ai ̄’vbv1⁄4      3 + 1 2  – 1  (2 – 1) GLv‡b, cÖavb Aÿ x A‡ÿi mgvšÍivj Dc‡K›`a؇qi `~iZ¡, 2ae = 2  ae = 1 ...... (i) cÖkœg‡Z, (a – ae)(a + ae) = 4  a 2 – 1 2 = 4  a 2 = 5 Avgiv Rvwb, a 2 e 2 = a2 – b 2  b 2 = a2 – a 2 e 2 = 5 – 1 = 4  Dce„‡Ëi mgxKiY: (x – 2) 2 ( 5) 2 + (y + 1) 2 2 2 = 1  (x – 2) 2 5 + (y + 1) 2 4 = 1 (Ans.) 6. (1, 2) Dc‡K›`a, 2 Dr‡Kw›`aKZv Ges 2x + y = 1 wØKvÿwewkó KwY‡Ki mgxKiY wbY©q Ki| [BUET 20-21] mgvavb: KwY‡Ki mgxKiY,  SP = e.PM  (x – 1) 2 + (y – 2) 2 = 2     2x + y – 1 2 2 + 12  x 2 – 2x + 1 + y2 – 4y + 4 = 2  (2x + y – 1) 2 5  5x2 + 5y2 – 10x – 20y + 25 = 2(4x2 + y2 + 1 + 4xy – 2y – 4x)  3x2 – 3y2 + 8xy + 2x + 16y – 23 = 0 (Ans.) 7. GKwU Dce„‡Ëi mgxKiY wbY©q Ki, hvi Dc‡K‡›`ai ̄’vbv1⁄4 (0, 2), Dr‡Kw›`aKZv 1 2 Ges wbqvgK‡iLvi mgxKiY y + 4 = 0| Gi Dc‡Kw›`aK j‡¤^i •`N© ̈I wbY©q Ki| [BUET 19-20] mgvavb: (x – 0) 2 + (y – 2) 2 = 1 2     y + 4 1  x 2 + y2 + 4 – 4y = 1 4 (y2 + 16 + 8y)  4x2 + 4y2 – 16y + 16 = y2 + 8y + 16  4x2 + 3y2 – 24y = 0  4x2 + 3(y2 – 8y + 42 ) = 48  x 2 12 + (y – 4) 2 4 2 = 1 (Ans.); GLv‡b, 4 2 > 12  Dc‡Kw›`aK j‡¤^i •`N© ̈ = 2a2 b = 2  12 4 = 6 GKK (Ans.) 8. GKwU Dce„‡Ëi mgxKiY wbY©q Ki, hvi GKwU Dc‡K‡›`ai ̄’vbv1⁄4 (1, – 1), Abyiƒc w`Kvÿ x – y – 4 = 0 Ges hv (1, 1) we›`y w`‡q AwZμg K‡i| [BUET 17-18] mgvavb: Dce„‡Ëi mgxKiY, (x – 1)2 + (y + 1)2 = e2 (x – y – 4) 2 2 ; hv (1, 1) we›`yMvgx|  (1 – 1)2 + (1 + 1)2 = e2 (1 – 1 – 4) 2 2  e 2 = 1 2  e = 1 2  mgxKiYwU, (x – 1)2 + (y + 1)2 = 1 2  (x – y – 4) 2 2  4x2 + 4y2 – 8x + 8y + 8 = x2 + y2 + 16 – 2xy – 8x + 8y  3x2 + 3y2 + 2xy – 8 = 0 (Ans.) 9. Ggb GKwU cive„‡Ëi mgxKiY wbY©q Ki hvi kxl©we›`y(4, – 3), Dc‡Kw›`aK j‡¤^i •`N© ̈ 4 Ges hvi Aÿ x A‡ÿi mgvšÍivj| [BUET 13-14; KUET 10-11; RUET 05-06; BUTex 04-05] mgvavb: GLv‡b, Dc‡Kw›`aK j‡¤^i •`N© ̈, |4a| = 4  4a =  4  Aÿ‡iLv x A‡ÿi mgvšÍivj Ges (4, – 3) kxl© wewkó cive„‡Ëi mgxKiY: (y + 3)2 =  4(x – 4) (Ans.) 10. GKwU Dce„‡Ëi AÿØq ̄’vbv‡1⁄4i AÿØq eivei Aew ̄’Z| Dce„ËwU x 9 + y 4 = 1 †iLv‡K x A‡ÿi Dci Ges x 2 + y 3 = 1 †iLv‡K y A‡ÿi Dci †Q` K‡i| Dce„ËwUi mgxKiY, Dr‡Kw›`aKZv Ges Dc‡K›`a `ywUi ̄’vbv1⁄4 wbY©q Ki| [BUET 11-12] mgvavb: Dce„‡Ëi, x A‡ÿi †Q`we›`y (9, 0)  a = 9 y A‡ÿi †Q`we›`y (0, 3)  b = 3  Dce„‡Ëi mgxKiY: x 2 9 2 + y 2 3 2 = 1 (Ans.)  Dr‡Kw›`aKZv, e = 1 – b 2 a 2 = 1 – 3 2 9 2 = 72 9 (Ans.)  Dc‡K›`a ( ae, 0)       9  72 9  0  ( 72 0) (Ans.) 11. 9x2 – 16y2 + 72x – 32y – 16 = 0 eμ‡iLvwUi cÖK...wZ, Zvi †K›`a I Dc‡K›`a؇qi ̄’vbv1⁄4, wbqvgK؇qi mgxKiY Ges bvwfj‡¤^i •`N© ̈ wbY©q Ki| [BUET 10-11] mgvavb: 9x2 – 16y2 + 72x – 32y – 16 = 0  9(x + 4)2 – 16(y + 1)2 = 144  (x + 4) 2 16 – (y + 1) 2 9 = 1 ...... (i)  eμ‡iLvwU Awae„Ë| (Ans.) †K›`a n‡jv, (x + 4, y + 1) = 0  (x, y)  (– 4, – 1) (Ans.)  e = 16 + 9 16 = 5 4