Tiêu đề Conics Engineering Practice Sheet Solution.pdf ✅

KwYK  Engineering Practice Sheet Solution 1 06 KwYK Conics WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. 4x2 – 8x – 5y2 – 20y – 36 = 0 mgxKiYwU Av`k© AvKv‡i cÖKvk Ki Ges Gi bvwfj‡¤^i mgxKiY †ei Ki| [BUET 23-24] mgvavb: 4x2 – 8x – 5y2 – 20y – 36 = 0  4(x2 – 2x + 1) – 5(y2 + 4y + 4) = 36 – 20 + 4  4(x – 1)2 – 5(y + 2)2 = 20  (x – 1) 2 5 – (y + 2) 2 4 = 1  Dr‡Kw›`aKZv, e = 1 + 4 5 = 3 5 a = 5  bvwfj‡¤^i mgxKiY: x – 1 =  ae  x – 1 =  5  3 5  x – 1 =  3 (Ans.) 2. GKwU MÖ‡ni Dce„ËvKvi Kÿc‡_i `ywU kxl©we›`yi `~iZ¡ 1400 wgwjqb wK‡jvwgUvi Ges `ywU †dvKvm we›`yi ga ̈eZ©x `~iZ¡ 800 (eight hundred) wgwjqb wK‡jvwgUvi| Dce„ËvKvi Kÿc_wUi mgxKiY wbY©q Ki| [BUET 22-23] mgvavb: 1400 800 GLv‡b, 2a = 1400  a = 700 Ges 2ae = 800  ae = 400 GLb, a 2 e 2 = a2 – b 2  b 2 = a2 – a 2 e 2 = 7002 – 4002 = 330000  wb‡Y©q mgxKiY: x 2 a 2 + y 2 b 2 = 1  x 2 490000 + y 2 330000 = 1 (Ans.) 3. GKwU cive„ËvKvi DcMÖn wW‡ki e ̈vm 6 dzU I MfxiZv 2.5 dzU| hw` wiwmfviwU cive„‡Ëi †dvKv‡m ̄’vcb Kiv nq Zvn‡j kxl©we›`y n‡Z wiwmfviwUi `~iZ¡ KZ? [BUET 22-23] mgvavb: a 2.5ft 6ft (2.5, 3) (0, 0) (a, 0) (2.5, – 3) 3ft 3ft X Y awi, cive„‡Ëi mgxKiY: y 2 = 4ax  3 2 = 4a  2.5 [hv (2.5, 3) we›`yMvgx]  a = 9 10  wiwmfviwU kxl© n‡Z 9 10 ft `~i‡Z¡ Aew ̄’Z| (Ans.) 4. Aa©-Dce„ËvKvi GKwU Uv‡b‡ji cÖ ̄’ 36 ft, D”PZv 18 ft, GB Uv‡b‡ji wfZi w`‡q 12 ft cÖ‡ ̄’i GKwU UavK †h‡Z cvi‡j Uav‡Ki m‡e©v”P D”PZv KZ? [BUET 21-22] mgvavb: (0, 0) a 36 ft 6 ft 6 ft (– 6, y) (6, y) Uv‡b‡ji cÖ ̄’, 2a = 36 ft  a = 18 ft D”PZv, b = 18 ft  Dce„‡Ëi mgxKiY: x 2 182 + y 2 182 = 1  6 2 182 + y 2 182 = 1 [hv (6, y) we›`yMvgx]  y 2 182 = 1 – 36 324  y = 16.97 ft (cÖvq)  Uav‡Ki m‡e©v”P D”PZv 16.97 ft (cÖvq) (Ans.)
2  Higher Math 2 nd 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
KwYK  Engineering Practice Sheet Solution 3 Dc‡K›`a, (X, Y)       4  5 4  0  (x + 4, y + 1)  ( 5, 0)  (x, y)  (1, – 1), (– 9, – 1) (Ans.)  wbqvgK؇qi mgxKiY: X =  a e =  4 5 4  x + 4 =  16 5 (Ans.)  bvwfj‡¤^i •`N© ̈ = 2  9 4 = 9 2 GKK (Ans.) 12. †Kv‡bv Dce„‡Ëi ÿz`a A‡ÿi •`N© ̈ Zvi Dc‡K›`a `ywUi ga ̈Kvi `~i‡Z¡i mgvb Ges Dc‡Kw›`aK j¤^ 10 GKK, Dce„ËwUi Dr‡Kw›`aKZv I mgxKiY wbY©q Ki| [BUET 09-10] mgvavb: awi, Dce„‡Ëi mgxKiY: x 2 a 2 + y 2 b 2 = 1; a > b kZ©g‡Z, 2b = 2ae  b = ae  a 2 e 2 = a2 – b 2  b 2 = a2 – b 2  2b2 = a2 Avevi, 2b2 a = 10  a 2 a = 10  a = 10  b 2 = 100 2 = 50  e = 1 – 50 100 = 1 2 (Ans.) Ges mgxKiY: x 2 100 + y 2 50 = 1 (Ans.) 13. x – y + 2 = 0 †iLvwU †Kv‡bv cive„‡Ëi kxl©we›`y‡Z Zvi A‡ÿi Dci j¤^| cive„‡Ëi †dvKvm (1, – 1) we›`y‡Z n‡j, Zvi mgxKiY wbY©q Ki| [BUET 08-09] mgvavb: kxl©Mvgx j¤^ x – y + 2 = 0 Ges †dvKvm (1, – 1)  Aÿ‡iLv x + y = 0 Ges kxl© ( – 1, 1) awi, w`Kv‡ÿi cv`we›`y (   + 1 2 = – 1,   – 1 2 = 1  = – 3,   = 3  (, )  (– 3, 3)  w`Kv‡ÿi mgxKiY: x – y = – 3 – 3  x – y + 6 = 0  cive„‡Ëi mgxKiY, (x – 1)2 + (y + 1)2 = (x – y + 6) 2 ( 1 ) 2 + (– 1) 2 2  x 2 – 2x + 1 + y2 + 2y + 1 = x 2 + y2 + 36 – 2xy – 12y + 12x 2  2x2 – 4x + 2 + 2y2 + 4y + 2 = x2 + y2 + 36 – 2xy – 12y + 12x  x 2 + y2 + 2xy = 16x – 16y + 32  (x + y)2 = 16(x – y + 2) (Ans.) 14. GKwU cive„‡Ëi mgxKiY wbY©q Ki, hvi Dc‡K›`a g~j we›`y‡Z Aew ̄’Z Ges x – y + 1 = 0 †iLvwU cive„ˇK Gi kxl©we›`y‡Z ̄úk© K‡i| [BUET 05-06; KUET 11-12] mgvavb: kxl©Mvgx j¤^ x – y + 1 = 0 Ges †dvKvm (0, 0)  Aÿ‡iLv x + y = 0 Ges kxl©    –  1 2  1 2 awi, w`Kv‡ÿi cv`we›`y (   + 0 2 = – 1 2 ,   + 0 2 = 1 2  = – 1,   = 1  (, )  (– 1, 1)  w`Kv‡ÿi mgxKiY: x – y = – 1 – 1  x – y + 2 = 0  cive„‡Ëi mgxKiY, (x – 0)2 + (y – 0)2 = (x – y + 2) 2 ( 1 ) 2 + (– 1) 2 2  2x2 + 2y2 = x 2 + y2 + 4 – 2xy – 4y + 4x  (x + y)2 = 4(x – y + 1) (Ans.) 15. (3, 4) Dc‡K›`a I (0, 0) kxl©wewkó cive„‡Ëi w`Kv‡ÿi mgxKiY wbY©q Ki| [BUET 05-06] mgvavb: wbqvg‡Ki cv`we›`y (, ) n‡j,  + 3 2 = 0,  + 4 2 = 0  (, )  (– 3, – 4) GLb, A‡ÿi Xvj = 4 3  w`Kv‡ÿi Xvj = – 3 4 Z( A(0, 0) S(3, 4) , )  w`Kv‡ÿi mgxKiY: y + 4 = – 3 4 (x + 3)  3x + 4y + 25 = 0 (Ans.) 16. 5x2 + 4y2 = 1 Dce„‡Ëi w`Kvÿ ̧‡jvi mgxKiY I †dvKvmØq wK wK? [BUET 04-05] mgvavb: mgxKiYwU, x 2 1 5 + y 2 1 4 = 1; b > a Dr‡Kw›`aKZv, e = 1 4 – 1 5 1 4 = 1 5  w`Kvÿ, y =  b e  y =  5 2  2y  5 = 0 (Ans.)  †dvKvm (0,  be)      0  1 2 5 (Ans.)
4  Higher Math 2 nd Paper Chapter-6 17. x 2 + 4x + 2y = 0 cive„‡Ëi kxl©we›`y I w`Kv‡ÿi mgxKiY wbY©q Ki| [BUET 04-05] mgvavb: x 2 + 4x + 2y = 0  (x + 2)2 = – 2(y – 2)  (x + 2)2 = 4   –  1 2 (y – 2) kxl©we›`y, x + 2 = 0, y – 2 = 0  x = – 2, y = 2  kxl©we›`y  (– 2, 2) (Ans.) Avevi, w`Kvÿ †iLvi mgxKiY, y – 2 = –    –  1 2  y = 5 2 (Ans.) 18. x + y = a †K `yBevi eM© K‡i KwYKwU mbv3 Ki| A‡ÿi mgxKiY, kxl©we›`y Ges ̄’vbv1⁄4 Aÿ؇qi ̄úk© we›`y †`wL‡q Qwe AuvK| [BUET 02-03] mgvavb: x + y = a  x = a – y  x = a + y – 2 ay [eM© K‡i]  (x – y – a)2 = 4ay [cybivq eM© K‡i]  x 2 + y2 + a2 – 2xy + 2ay – 2ax = 4ay  (x – y)2 = 2ay + 2ax – a 2  (x – y)2 = a(2y + 2x – a)  KwYKwU GKwU cive„Ë| y = x (a, 0) (0, a) ( ) a 4  a 4 Y X O  A‡ÿi mgxKiY: x – y = 0 kxl©we›`y n‡e, x – y = 0 Ges x + y = a Gi †Q`K|  kxl©we›`yi ̄’vbv1⁄4     a 4  a 4 , x A‡ÿ †Q`we›`y(a, 0), y A‡ÿ †Q`we›`y (0, a) weMZ mv‡j KUET-G Avmv cÖkœvejx 19. (2sec, 3tan) we›`yi mÂvic_ wbY©q Ki Ges †mLvb †_‡K H KwY‡Ki w`Kvÿ, Dr‡Kw›`aKZv I Dc‡K›`a wbY©q Ki| [KUET 19-20] mgvavb: (x, y)  (2sec, 3tan)  sec = x 2 Ges tan = y 3 sec2  – tan2  = 1  x 2 2 2 – y 2 3 2 = 1; hv GKwU Awae„Ë| (Ans.)  Dr‡Kw›`aKZv, e = 1 + b 2 a 2 = 1 + 3 2 2 2 = 13 2 (Ans.)  w`Kvÿ †iLvi mgxKiY: x =  a e =  2 13 2  x =  4 13  13x =  4 (Ans.)  Dc‡K‡›`ai ̄’vbv1⁄4 ( ae, 0)       2  13 2  0  ( 13 0) (Ans.) 20. GKwU cive„‡Ëi mgxKiY wbY©q Ki, hvi Dc‡K›`a g~j we›`y‡Z Aew ̄’Z Ges x – y + 1 = 0 †iLvwU cive„ˇK Gi kxl©we›`y‡Z ̄úk© K‡i| [KUET 11-12; BUET 05-06] mgvavb: kxl©Mvgx j¤^ x – y + 1 = 0 Ges †dvKvm (0, 0)  Aÿ‡iLv x + y = 0 Ges kxl©    –  1 2  1 2 awi, w`Kv‡ÿi cv`we›`y (   + 0 2 = – 1 2 ,   + 0 2 = 1 2  = – 1,   = 1  (, )  (– 1, 1)  w`Kv‡ÿi mgxKiY: x – y = – 1 – 1  x – y + 2 = 0  cive„‡Ëi mgxKiY, (x – 0)2 + (y – 0)2 = (x – y + 2) 2 ( 1 ) 2 + (– 1) 2 2  2x2 + 2y2 = x 2 + y2 + 4 – 2xy – 4y + 4x  (x + y)2 = 4(x – y + 1) (Ans.) 21. 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| [KUET 10-11; BUET 13-14; 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.) 22. (– 1, 1) Dc‡K›`a Ges (2, – 3) kxl©we›`y wewkó cive„ËwUi Aÿ I wbqvg‡Ki mgxKiY wbY©q Ki| [KUET 05-06] mgvavb: A‡ÿi mgxKiY: y – 1 x + 1 = 1 + 3 – –   – 3y + 3 = 4x + 4  4x + 3y + 1 = 0 (Ans.) wbqvg‡Ki cv`we›`y (, ) n‡j,  – 1 2 = 2,  + 1 2 = – 3  (, )  (5, – 7)  wbqvg‡Ki mgxKiY: 3x – 4y = 15 + 28  3x – 4y – 43 = 0 (Ans.)