Tiêu đề Conics Varsity Practice Sheet Solution.pdf ✅

KwYK  Varsity Practice Sheet Solution 1 06 KwYK Conics weMZ mv‡j DU-G Avmv cÖkœvejx 1. 16y2 – 9x2 + 18x + 64y + 199 = 0 Øviv wb‡`©wkZ KwYK †KvbwU? [DU 23-24] Awae„Ë (hyperbola) e„Ë (circle) cive„Ë (parabola) Dce„Ë (ellipse) DËi: Awae„Ë (hyperbola) e ̈vL ̈v: a = – 9, b = 16, h = 0  ab – h 2 = – 144 < 0  GwU Awae„Ë| 2. y 2 = 8x + 2y – 9 cive„‡Ëi Dc‡K‡›`ai ̄’vbv1⁄4 †KvbwU? [DU 21-22] (3, 1) (3, 0) (– 1, 1) (2, 0) DËi: (3, 1) e ̈vL ̈v: y 2 = 8x + 2y – 9  y 2 – 2y = 8x – 9  y 2 – 2y + 1 = 8x – 9 + 1  (y – 1)2 = 8x – 8  (y – 1)2 = 8(x – 1)  (y – 1)2 = 4  2(x – 1)  a = 2 I kxl© (1, 1) GLb, x – 1 = a Ges y – 1 = 0  x – 1 = 2  y = 1  x = 3  Dc‡K‡›`ai ̄’vbv1⁄4 (3, 1) 3. x 2 – 3y2 – 2x = 8 Awae„‡Ëi kxl©we›`yØq- [DU 21-22] (– 4, 0), (2, 0) (0, 4), (0, 2) (0, – 4), (0, – 1) (4, 0), (– 2, 0) DËi: (4, 0), (– 2, 0) e ̈vL ̈v: x 2 – 2x + 1 – 3y2 = 9  (x – 1)2 – 3y2 = 9  (x – 1) 2 9 – y 2 3 = 1  a 2 = 9 Ges b 2 = 3  a = 3  b = 3 GLb, x – 1 =  3  x = 4, – 2  kxl©we›`yØq (4, 0), (– 2, 0) 4. GKwU Dce„‡Ëi Dc‡Kw›`aK j‡¤^i •`N© ̈ e„n`v‡ÿi •`‡N ̈©i A‡a©K, Dce„ËwUi Dr‡Kw›`aKZv KZ? [DU 21-22] 1 3 1 2 2 3 2 3 DËi: 1 2 e ̈vL ̈v: Dc‡Kw›`aK j‡¤^i •`N© ̈ = 2b2 a [†hLv‡b, a > b] e„n`vÿ = 2a  2b2 a = 1 2  2a  b 2 a 2 = 1 2  e = 1 – b 2 a 2 = 1 – 1 2 = 1 2 5. 2x = y2 + 8y + 22 cive„‡Ëi kxl©we›`yi ̄’vbv1⁄4 n‡eÑ [DU 20-21] (3, – 4) (– 3, 4) (– 3, – 4) (3, 4) DËi: (3, – 4) e ̈vL ̈v: 2x = y2 + 8y + 22  2x = y2 + 2  4y + 16 + 6  (y + 4)2 = 2x – 6  (y + 4)2 = 2(x – 3)  kxl©we›`y(3, – 4) 6. 25x2 + 16y2 = 400 Dce„‡Ëi Dr‡Kw›`aKZv KZ? [DU 19-20, 17-18] 2 3 4 5 3 4 3 5 DËi: 3 5 e ̈vL ̈v: x 2 16 + y 2 25 = 1  a 2 = 16 Ges b 2 = 25  e = 1 – a 2 b 2 = 1 – 16 25 = 3 5
2  Higher Math 2 nd Paper Chapter-6 7. y 2 – 4y – x 2 + 6x = 12 mgxKiYwU †Kvb ai‡bi KwYK? [DU 18-19] e„Ë (Circle) Dce„Ë (Ellipse) cive„Ë (Parabola) Awae„Ë (Hyperbola) DËi: Awae„Ë (Hyperbola) e ̈vL ̈v: a = – 1, b = 1, h = 0  ab – h 2 = – 1 – 0 = – 1 < 0  GwU Awae„Ë| 8. 2x2 – 8y2 = 2 Awae„‡Ëi Dr‡Kw›`aKZvi gvb- [DU 18-19] 3 2 2 3 2 5 2 5 2 DËi: 5 2 e ̈vL ̈v: x 2 1 – y 2 1 4 = 1  a 2 = 1 Ges b 2 = 1 4  e = 1 + b 2 a 2 = 1 + 1 4 = 5 2 9. (4, 3) we›`y‡Z 3x2 – 4y2 = 12 Awae„‡Ëi ̄úk©‡Ki Xv‡ji gvb- [DU 17-18] – 1 1 3 4 4 3 DËi: 1 e ̈vL ̈v: 3x2 – 4y2 = 12  6x – 4  2y dy dx = 0 [x Gi mv‡c‡ÿ AšÍixKiY K‡i]  8y dy dx = 6x  dy dx = 6x 8y = 3x 4y  (4, 3) we›`y‡Z ̄úk©‡Ki Xvj = 3x 4y = 3  4 4  3 = 1 10. y 2 + 4x + 2y – 8 = 0 cive„‡Ëi kxl©we›`y n‡eÑ [DU 16-17]     9 4  – 1    –  9 4  1 (0, 1) (2, 0) DËi:     9 4  – 1 e ̈vL ̈v: y 2 + 4x + 2y – 8 = 0  y 2 + 2y + 1 + 4x – 9 = 0  (y + 1)2 = – 4x + 9  (y + 1)2 = – 4     x – 9 4  kxl©we›`y     9 4  – 1 11. 4x2 + y2 = 2 Dce„ËwUi e„nr I ÿz`a A‡ÿi •`N© ̈ h_vμ‡g- [DU 16-17] 4 I 2 2 I 4 2 I 2 2 2 2 I 2 DËi: 2 2 I 2 e ̈vL ̈v: x 2 1 2 + y 2 2 = 1; a = 1 2 Ges b = 2 ⸪ b > a  e„nr A‡ÿi •`N© ̈ = 2b = 2 2 GKK ÿz`a A‡ÿi •`N© ̈ = 2a = 2 2 = 2 GKK 12. 5x2 + 15x – 10y – 4 = 0 cive„‡Ëi wbqvg‡Ki mgxKiYÑ [DU 14-15] 40x + 81 = 0 2x + 3 = 0 40y + 81 = 0 40y + 41 = 0 DËi: 40y + 81 = 0 e ̈vL ̈v: 5x2 + 15x – 10y – 4 = 0  5(x2 + 3x) – 10y – 4 = 0  5     x 2 + 2  3 2 x + 9 4 – 10y – 4 – 5  9 4 =0  5     x + 3 2 2 = 10y + 61 4  5     x + 3 2 2 = 10    y + 61 40      x + 3 2 2 = 2    y + 61 40  a = 1 2  wbqvg‡Ki mgxKiY: y + 61 40 = – a  y + 61 40 = – 1 2  y + 61 40 + 1 2 = 0  40y + 81 = 0 13. x 2 – 4x + 12y – 40 = 0 cive„‡Ëi Dc‡Kw›`aK j‡¤^i •`N© ̈ KZ? [DU 13-14; JU 18-19] 4 6 8 12 DËi: 12
KwYK  Varsity Practice Sheet Solution 3 e ̈vL ̈v: x 2 – 4x + 12y – 40 = 0  x 2 – 2  2x + 4 + 12y – 44 = 0  x 2 – 2  2x + 22 = – 12y + 44  (x – 2)2 = – 12   y –  11 3  (x – 2)2 = 4(– 3)   y –  11 3  a = – 3  Dc‡Kw›`aK j‡¤^i •`N© ̈ = |4a| = 12 GKK weMZ mv‡j GST-G Avmv cÖkœvejx 1. 2x2 + y2 – 8x – 2y + 1 = 0 Dce„ËwUi †K‡›`ai ̄’vbv1⁄4 †KvbwU? [GST 22-23] (2, 1) (– 2, 1) (1, 2) (1, – 2) DËi: (2, 1) e ̈vL ̈v: 2(x2 – 4x + 4) + y2 – 2y + 1 – 8 = 0  2(x – 2) 2 8 + (y – 1) 2 8 = 1  (x – 2) 2 4 + (y – 1) 2 8 = 1  †K›`a (2, 1) 2. x 2 – 8x + 4y – 4 = 0 KwYKwUi w`Kv‡ÿi cv`we›`yi ̄’vbv1⁄4Ñ [GST 22-23] (4, 6) (4, – 6) (– 4, – 6) (6, 4) DËi: (4, 6) e ̈vL ̈v: x 2 – 8x + 16 + 4y – 4 – 16 = 0  x 2 – 2  4x + 42 + 4y – 20 = 0  (x – 4)2 = – 4y + 20  (x – 4)2 = – 4(y – 5)  kxl© (4, 5) I a = – 1 GLb, x – 4 = 0 Ges y – 5 = 1  x = 4  y = 6  w`Kv‡ÿi cv`we›`y (4, 6) Note: (y – ) 2 = 4a(x – ) cive„‡Ëi w`Kv‡ÿi cv`we›`y( – a, ) (x – ) 2 = 4a(y – ) cive„‡Ëi w`Kv‡ÿi cv`we›`y(,  – a) 3. x 2 30 + y 2 14 = 1 Dce„‡Ëi wbqvgK †iLv؇qi ga ̈eZx© `~iZ¡ KZ GKK? [GST 22-23] 7 14 15 30 DËi: 15 e ̈vL ̈v: a 2 = 30 Ges b 2 = 14  e = 1 – b 2 a 2 = 2 30 15 `~iZ¡ = 2a e = 2 30 2 30 15 = 15 GKK Note: x 2 a 2 + y 2 b 2 = 1 Dce„‡Ëi, a > b n‡j, (i) Dc‡K›`a؇qi ga ̈eZ©x `~iZ¡ = 2ae (ii) wbqvgK؇qi ga ̈eZ©x `~iZ¡ = 2a e b > a n‡j, (i) Dc‡K›`a؇qi ga ̈eZ©x `~iZ¡ = 2be (ii) wbqvgK؇qi ga ̈eZ©x `~iZ¡ = 2b e 4. †h KwY‡Ki civwgwZK mgxKiY x = 3 + at2 , y = 2at †mUvi kxl©we›`yi ̄’vbv1⁄4Ñ [GST 22-23] (0, 0) (2, 0) (3, 0) (2, 3) DËi: (3, 0) e ̈vL ̈v: x – 3 = at2  x – 3 a = t2 Ges y 2a = t  y 2 4a2 = t2  y 2 4a2 = x – 3 a  y 2 = 4a(x – 3)  kxl©we›`y (3, 0) 5. 2x2 + 3y2 – 12x + 12y + 29 = 0 KwYKwUi Dc‡Kw›`aK j‡¤^i •`N© ̈ KZ? [GST 21-22] 2 3 3 2 2 2 3 2 2 3 DËi: 2 2 3 e ̈vL ̈v: 2(x2 – 6x + 9) + 3(y2 + 4y + 4) – 30 + 29 = 0  2(x – 3)2 + 3(y + 2)2 = 1  a = 1 2 Ges b = 1 3  Dc‡Kw›`aK j‡¤^i •`N© ̈ = 2b2 a = 2 3 1 2 = 2 2 3 GKK 6. r(1 + cos) = 2 Gi Kv‡Z©mxq mgxKiYÑ [GST 21-22] x 2 + y2 + x – 2 = 0 y 2 – 4x = 4 x 2 + 4x = 2 y 2 + 4x = 4 DËi: y 2 + 4x = 4 e ̈vL ̈v: r + r cos = 2  x 2 + y2 = 2 – x  x 2 + y2 = 4 – 4x + x2  y 2 + 4x = 4
4  Higher Math 2 nd Paper Chapter-6 7. y 2 = 3x Ges x 2 = 3y cive„Ë؇qi †Q`we›`y wb‡q MgbKvix mij‡iLvi Xvj KZ? [GST 21-22] – 1 1 2 3 DËi: 1 e ̈vL ̈v: y 2 = 3x .......... (i) x 2 = 3y .......... (ii) x = y 2 3 (ii) bs G ewm‡q cvB,     y 2 3 2 = 3y  y 4 = 27y  y(y3 – 27) = 0  y = 0, 3 Ges x = 0, 3 [(i) bs G y Gi gvb ewm‡q]  †Q`we›`ymg~n (0, 0) Ges (3, 3)  wb‡Y©q mij‡iLvi Xvj = 3 – 0 3 – 0 = 1 8. (y + 3) 2 = 8(x + 3) cive„‡Ëi Dc‡K‡›`ai †cvjvi ̄’vbv1⁄4 †KvbwU? [GST 20-21]     2 –  3     2 3  3     2 4 3     2 3 –  6 DËi:     2 4 3 e ̈vL ̈v: (y + 3) 2 = 4  2(x + 3) GLb, x + 3 = 2 Ges y + 3 = 0  x = – 1  y = – 3  Dc‡K›`a (– 1 – 3)  r = 3 + 1 = 2   =  + tan–1 3 =  +  3 = 4 3 †cvjvi ̄’vbv1⁄4     2 4 3 weMZ mv‡j Agri-G Avmv cÖkœvejx 1. y = 2x + c †iLvwU x 2 4 + y 2 3 = 1 Dce„‡Ëi ̄úk©K n‡j, c Gi gvb KZ? [Agri. Guccho 21-22] 7 25 19 25 DËi: 19 e ̈vL ̈v: a 2 = 4 Ges b 2 = 3 Ges y = 2x + c Xvj, m = 2 Dce„‡Ëi ̄úk©‡Ki mgxKi‡Yi kZ©vbyhvqx, c 2 = b2 + a2m 2  c 2 = 3 + 4  2 2  c 2 = 19  c =  19 Shortcut: (i) y = mx + c †iLvwU, x 2 a 2 + y 2 b 2 = 1 Dce„‡Ëi ̄úk©K n‡j, c 2 = b2 + a2m 2 x 2 a 2 – y 2 b 2 = 1 Awae„‡Ëi ̄úk©K n‡j, c 2 = a2m 2 – b 2 y 2 b 2 – x 2 a 2 = 1 Awae„‡Ëi ̄úk©K n‡j, c 2 = b2 – a 2m 2 (ii) x 2 a 2 + y 2 b 2 = 1 Dce„‡Ëi (x1, y1) we›`y‡Z ̄úk©‡Ki mgxKiY, xx1 a 2 + yy1 b 2 = 1 x 2 a 2 – y 2 b 2 = 1 Awae„‡Ëi (x1, y1) we›`y‡Z ̄úk©‡Ki mgxKiY, xx1 a 2 – yy1 b 2 = 1 y 2 b 2 – x 2 a 2 = 1 Awae„‡Ëi (x1, y1) we›`y‡Z ̄úk©‡Ki mgxKiY, yy1 b 2 – xx1 a 2 = 1 2. 5y2 – 2x = 0 cive„‡Ëi Dc‡K›`a †KvbwU? [Agri. Guccho 19-20]     0  1 2     0  1 5     1 10  0     0  1 10 DËi:     1 10  0 e ̈vL ̈v: 5y2 = 2x  y 2 = 2 5 x  y 2 = 4  2 20 x  a = 1 10  Dc‡K›`a     1 10  0 Note: y 2 = 4ax cive„‡Ëi Dc‡K›`a (a, 0) x 2 = 4ay cive„‡Ëi Dc‡K›`a (0, a) weMZ mv‡j JU-G Avmv cÖkœvejx 1. y 2 = 4x cive„‡Ëi Dcwiw ̄’Z p we›`yi †KvwU 6 n‡j, H we›`yi Dc‡Kw›`aK `~iZ¡ KZ? [JU 22-23] 3 2 2 3 10 3 5 DËi: 10 e ̈vL ̈v: 6 2 = 4x  x = 9  p we›`y (9, 6), a = 1  Dc‡Kw›`aK `~iZ¡ = a + x = 1 + 9 = 10 GKK Note: y 2 = 4ax cive„‡Ëi (x, y) we›`yi Dc‡Kw›`aK `~iZ¡ = a + x Ges x 2 = 4ay cive„‡Ëi (x, y) we›`yi Dc‡Kw›`aK `~iZ¡ = a + y