Tiêu đề Conics Practice Sheet Solution HSC FRB 25.pdf ✅

KwYK  Final Revision Batch '25 1 Board Questions Analysis m„Rbkxj cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 2 1 2 2 2 2 2 2 1 2022 2 2 2 2 1 2 2 2 2 eûwbe©vPwb cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 5 5 5 5 3 3 5 5 5 2022 5 5 5 5 5 5 5 6 5 weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| DÏxcK-1: 16x2 + 25y2 – 32x + 100y – 284 = 0 DÏxcK-2: Y Y X X C B P(6, 0) A(0, 4) O(0, 0) OC = 3 [XvKv †evW©- Õ23] (K) 4x2 – 9y2 = 36 Awae„‡Ëi AmxgZ‡Ui mgxKiY wbY©q Ki| (L) DÏxcK-1 Gi wbqvgK †iLvi mgxKiY wbY©q Ki| (M) DÏxcK-2 Gi wPÎwU GKwU cive„Ë Ges kxl©we›`y A n‡j, CB †iLvi •`N© ̈ wbY©q Ki| mgvavb: (K) GLv‡b, 4x2 – 9y2 = 36  4x2 36 – 9y2 36 = 1  x 2 9 – y 2 4 = 1  x 2 3 2 – y 2 2 2 = 1 ........ (i)  a = 3, b = 2  Awae„‡Ëi AmxgZ‡Ui mgxKiY, y =  b a x  y =  2 3 x (Ans.) (L) DÏxcK-1 G, 16x2 + 25y2 – 32x + 100y – 284 = 0  16(x2 – 2x) + 25(y2 + 4y) – 284 = 0  16(x2 – 2x + 1) + 25(y2 + 4y + 4) – 284 – 16 – 100 = 0  16(x – 1)2 + 25 (y + 2)2 = 400  (x – 1) 2 25 + (y+ 2) 2 16 = 1  (x – 1) 2 5 2 + (y+ 2) 2 4 2 = 1....... (i)  a = 5 Ges b = 4  a > b  Dr‡Kw›`aKZv, e = 1 – b 2 a 2 = 1 – 16 25 = 9 25 = 3 5 wbqvgK †iLvi mgxKiY, x – 1 =  5 3 5  x – 1 =  25 3 (+) wPý wb‡q cvB, x – 1 = 25 3  x = 25 3 + 1  x = 28 3  3x – 28 = 0 (–) wPý wb‡q cvB, x – 1 = – 25 3  x = 1 – 25 3  x = – 22 3  3x + 22 = 0  wbqvgK‡iLvi mgxKiY, 3x – 28 = 0, 3x + 22 = 0

KwYK  Final Revision Batch '25 3  y 2 ( 5) 2 – x 2 ( 3) 2 = 1  b = 5 ; a = 3 Ges b > a  Dr‡Kw›`aKZv, e = 1 + a 2 b 2 = 1 + 3 5 = 2 2 5 Dc‡K›`a (0,  be)        0   5 . 2 2 5  (0   2 2) (Ans.) (L) `„k ̈Kí-1 n‡Z cvB, 3x2 + 9x – 6y – 8 = 0 ...... (i)  3(x2 + 3) – 6y – 8 = 0  3       x 2 + 2x 3 2 +     3 2 2 – 6y – 8 – 3 9 4 = 0  3     x + 3 2 2 = 6y + 59 4  3     x + 3 2 2 = 6    y + 59 24      x + 3 2 2 = 2    y + 59 24 hv GKwU cive„‡Ëi mgxKiY| G‡K X 2 = 4aY Gi mv‡_ Zzjbv K‡i cvB, X = x + 3 2 , Y = y + 59 24 , 4a = 2  a = 1 2 GLb, Dc‡Kw›`aK j‡¤^i cÖvšÍ we›`y؇qi ̄’vbvsK, (X, Y) = ( 2a, a)  X =  2a  x + 3 2 =  2  1 2  x + 3 2 =  1  x = – 3 2  1  x = – 1 2 , – 5 2 Ges Y = a  y + 59 24 = 1 2  y = 1 2 – 59 24  y = – 47 24  Dc‡Kw›`aK j‡¤^i cÖvšÍwe›`y؇qi ̄’vbv1⁄4    –  5 2  – 47 24 I    –  1 2  – 47 24 (Ans.) wbqvgK‡iLvi mgxKiY, Y = – a A_©vr, y + 59 24 = – 1 2  y = – 1 2 – 59 24 = – 12 – 59 24 = – 71 24 = – 71 24  24y + 71 = 0 (Ans.) (M) awi, KwYKwUi mgxKiY, x 2 a 2 + y 2 b 2 = 1 ..... (i) †hLv‡b, a > b †`Iqv Av‡Q, Dr‡Kw›`aKZv, e = 1 3 Ges Dc‡Kw›`aK j‡¤^i •`N© ̈, 10  2b2 a = 10  b 2 = 5a Avevi, e = 1 – b 2 a 2 [⸪ a > b]  1 3 = 1 – b 2 a 2  1 3 = 1 – 5a a 2 = 1 – 5 a  5 a = 1 – 1 3 = 2 3  a = 15 2  b 2 = 5.a = 5  15 2 = 75 2 (i) bs n‡Z cvB, x 2 a 2 + y 2 b 2 = 1  x 2     15 2 2 + y 2 75 2 = 1  4x2 225 + 2y2 75 = 1  4x2 + 6y2 = 225 (Ans.) 4| `„k ̈Kí-1: GKwU cive„‡Ëi kxl©we›`y (1, 1) Ges wbqvgK‡iLvi mgxKiY, 2x + y – 1 = 0 `„k ̈Kí-2: X S(–12, 6) A C A S(12, 6) X Y Y [ivRkvnx †evW©- Õ23] (K) 2x2 + y2 = 2 KwYKwUi kxl©we›`yi ̄’vbv1⁄4 wbY©q Ki| (L) `„k ̈Kí-1 Gi Av‡jv‡K cive„‡Ëi mgxKiY wbY©q Ki| (M) `„k ̈Kí-2 Gi Dr‡Kw›`aKZv 3 n‡j KwYKwUi AmxgZU †iLvi mgxKiY wbY©q Ki| mgvavb: (K) 2x2 + y2 = 2  x 2 1 + y 2 2 = 1  x 2 1 2 + y 2 ( 2) 2 = 1 ; hv GKwU Dce„‡Ëi mgxKiY|  a = 1 ; b = 2  b > a  kxl©we›`yi ̄’vbv1⁄4 (0,  b)  (0   2) (Ans.)