Tiêu đề Conics- Daily-10 MCQ (Set-A)-With Solve.pdf ✅

1 Varsity Daily-10 [Set-A (Solve Sheet)] wm‡jevm: KwbK c~Y©gvb: 30 †b‡MwUf gvK©: 0.25 mgq: 20 wgwbU 1. x 2 5 – y 2 4 = 1 Awae„‡Ëi wbqvg‡Ki mgxKiY †KvbwU? [Determine the equation of the directrix of the hyperbola : x 2 5 – y 2 4 = 1] 3x  5 = 0 x  2 = 0 2x + 70 = 0 3x + 2 = 0 DËi: 3x  5 = 0 e ̈vL ̈v: x 2 5 – y 2 4 = 1 Awae„‡Ëi a = 5, b = 2; †hLv‡b a > b myZivs wbqvgK †iLvi mgxKiY, x =  a e  x =  5 3 5  x =  5 3  3x  5 = 0 Dr‡Kw›`aKZv, e = 1 + b 2 a 2 = 1 + 4 5 = 3 5 2. y = 2x + c †iLvwU x 2 4 + y 2 3 = 1 Dce„‡Ëi ̄úk©K n‡j, c Gi gvb KZ? [If the line y = 2x + c is tangent to the ellipse x 2 4 + y 2 3 = 1, what is the value of c?] 19 25  19  13 DËi:  19 e ̈vL ̈v: y = mx + c †iLvwU x 2 a 2 + y 2 b 2 = 1 GB Dce„‡Ëi ̄úk©K n‡j, c 2 = a2m 2 + b2 †`Iqv Av‡Q, y = 2x + c †hLv‡b, m = 2 x 2 4 + y 2 3 = 1  x 2 2 2 + y 2 ( 3) 2 = 1; †hLv‡b, a = 2 Ges b = 3 myZivs, c 2 = (2)2  (2)2 + ( 3) 2 = 4  4 + 3 = 19 c =  19 3. Awae„‡Ëi civwgwZK ̄’vbv1⁄4 (5sec, 4tan) n‡j, Awae„‡Ëi mgxKiY †KvbwU? [If the parametric coordinates of a hyperbola are (5secθ, 4tanθ), what is the equation of the hyperbola?] 25y 2 – 16x2 = 400 25x 2 – 16y2 = 400 16x 2 – 25y2 = 400 4x 2 – 5y2 = 20 DËi: 16x 2 – 25y2 = 400 e ̈vL ̈v: x 2 a 2 – y 2 b 2 = 1 Awae„‡Ëi civwgwZK ̄’vbv1⁄4, x = asec y = btan x = 5sec, y = 4tan myZivs, a = 5 Ges b = 4  Awae„‡Ëi mgxKiY x 2 5 2 – y 2 4 2 = 1  16x2 – 25y2 = 400 [Dfqc‡ÿ 400 Øviv ̧Y K‡i] 4. 4x2 + 5y2 = 20 Dce„‡Ëi Dr‡Kw›`aKZv KZ? [What is the eccentricity of the ellipse 4x2 + 5y2 = 20?] 5 3 4 1 5 1 2 DËi: 1 5 e ̈vL ̈v: 4x2 + 5y2 = 20  x 2 5 + y 2 4 = 1; †hLv‡b a = 5 Ges b = 2; a > b  x 2 ( 5) 2 + y 2 (2) 2 = 1 Dr‡Kw›`aKZv, e = 1 – b 2 a 2 = 1 – 4 5 = 1 5 5. (x – 5) 2 100 + (y + 4) 2 64 = 1 Dce„‡Ëi †dvKvm؇qi `~iZ¡ KZ GKK? [What is the distance between the focus of the ellipse (x – 5) 2 100 + (y + 4) 2 64 = 1?] 6 12 3 24 DËi: 12 e ̈vL ̈v: (x – 5) 2 100 + (y + 4) 2 64 = 1  (x – 5) 2 102 + (y + 4) 2 8 2 = 1; †hLv‡b a = 10 Ges b = 8; a > b  †dvKvm؇qi `~iZ¡ = 2ae = 2 a 2 – b 2 = 2 100 – 64 = 2 36 = 12 GKK Dce„‡Ëi a > b n‡j, ae = a 2 – b 2 a < b n‡j, be = b 2 – a 2 6. 9x2 + 25y2 = 225 Dce„‡Ëi Dc‡Kw›`aK j‡¤^i mgxKiY Kx n‡e? [What is the equation of the latus rectum of the ellipse 9x2 + 25y2 = 225?] y =  4 x =  4 x =  5 y =  5 DËi: x =  4
2 e ̈vL ̈v: 9x2 + 25y2 = 225  x 2 25 + y 2 9 = 1  x 2 5 2 + y 2 3 2 = 1 [a = 5 Ges b = 3; a > b]  Dc‡Kw›`aK j‡¤^i mgxKiY, x =  ae [Dce„‡Ëi a > b n‡j, ae = a 2 – b 2 ]  x =  a 2 – b 2  x =  25 – 9  x =  4 7. x 2 25 – y 2 16 = 1 Awae„‡Ëi AbyeÜx A‡ÿi •`N© ̈ KZ GKK? [What is the length of the conjugate axis of the hyperbola x 2 25 – y 2 16 = 1?] 10 8 4 5 DËi: 8 e ̈vL ̈v: Awae„Ë, x 2 25 – y 2 16 = 1  x 2 5 2 – y 2 4 2 = 1 †hLv‡b, a = 5 Ges b = 4; a > b  AbyeÜx A‡ÿi •`N© ̈ = 2b = 2  4 = 8 GKK| 8. 16x2 – 25y2 = 400 Awae„‡Ëi AmxgZ‡Ui mgxKiY wb‡Pi †KvbwU? [What is the equation of the asymptotes of the hyperbola 16x2 - 25y2 = 400?] y =  5 4 x y =  4 5 x x =  4 5 y x =  5 4 y DËi: y =  4 5 x e ̈vL ̈v: 16x2 – 25y2 = 400  x 2 25 – y 2 16 = 1  x 2 5 2 – y 2 4 2 = 1; †hLv‡b, a = 5, b = 4  AmxgZ‡Ui mgxKiY, y =  b a x  y =  4 5 x 9. x 2 = 4y + 3 cive„‡Ëi kxl©we›`yi ̄’vbv1⁄4 KZ? [What are the coordinates of the vertex of the parabola x2 = 4y + 3?]    –  3 4  0     0  3 4     0  – 3 4     3 4  0 DËi:     0  – 3 4 e ̈vL ̈v: x 2 = 4y + 3  x 2 = 4    y + 3 4  x 2 = 4  1      y + 3 4 kxl©we›`yi ̄’vbv1⁄4     x  y + 3 4 = (0, 0) x = 0 Ges y + 3 4 = 0  y = – 3 4  kxl©we›`yi ̄’vbv1⁄4 (x, y) =     0  – 3 4 10. 3x2 + 5y2 = 15 Dce„‡Ëi Dr‡Kw›`aKZv KZ? [What is the eccentricity of the ellipse 3x2 + 5y2 = 15?] 5 2 3 5 5 3 2 5 DËi: 2 5 e ̈vL ̈v: 3x2 + 5y2 = 15  x 2 5 + y 2 3 = 1 x 2 ( 5) 2 + y 2 ( 3) 2 = 1; †hLv‡b, a = 5 Ges b = 3, [a > b] Dce„‡Ëi Dr‡Kw›`aKZv = 1 – b 2 a 2 = 1 – 3 5 = 2 5 11. x 2 = – 3y cive„‡Ëi Dc‡K‡›`ai ̄’vbv1⁄4 †KvbwU? [What are the coordinates of the focus of the parabola x2 = -3y?] (0, – 3)     0  3 4 (0, 3)     0  – 3 4 DËi:     0  – 3 4 e ̈vL ̈v: x 2 = – 3y  x 2 = 4      – 3 4 y; †hLv‡b, a = – 3 4 Avgiv Rvwb, x 2 = 4ay cive„‡Ëi Dc‡K‡›`ai ̄’vbv1⁄4 (0, a)  x 2 = – 3y cive„‡Ëi Dc‡K‡›`ai ̄’vbv1⁄4     0  – 3 4 12. y 2 = 9x cive„‡Ëi Dcwiw ̄’Z P we›`yi †KvwU 12 n‡j, H we›`yi Dc‡Kw›`aK `~iZ¡ KZ GKK? [If the ordinate of the point P on the parabola y2 = 9x is 12, what is the point's eccentric distance?] 20.25 18.25 10.50 9.50 DËi: 18.25 e ̈vL ̈v: y 2 = 9x Gi Dcwiw ̄’Z we›`y P hvi †KvwU 12|  9x = (12)2  x = 144 9  x = 16 y 2 = 9x  y 2 = 4  9 4  x  a = 9 4
3 Avgiv Rvwb, y 2 = 4ax cive„‡Ëi Dc‡Kw›`aK `~iZ¡ = x + a  Dc‡Kw›`aK `~iZ¡ = x + a = 16 + 9 4 = 16 + 2.25 = 18.25 GKK 13. x = at2 , y = 2at civwgwZK mgxKiY wb‡`©kK mÂvic_ GKwUÑ [The parametric equations x = at2, y = 2at represent a:] e„Ë (circle) cive„Ë (parabola) Dce„Ë (ellipse) Awae„Ë (hyperbola) DËi: cive„Ë (parabola) e ̈vL ̈v: y = 2at  t = y 2a Avevi, x = at2  x = a      y 2a 2  x = ay2 4a2  x = y 2 4a  4ax = y2 hv cive„‡Ëi mgxKiY| 14. 25x2 + 16y2 = 400 Dce„ËwUi Dc‡Kw›`aK j‡¤^i •`N ̈© KZ GKK? [What is the length of the latus rectum of the ellipse 25x2 + 16y2 = 400?] 50 4 32 25 32 5 8 5 DËi: 32 5 e ̈vL ̈v: 25x2 + 16y2 = 400  x 2 16 + y 2 25 = 400  x 2 4 2 + y 2 5 2 = 1; †hLv‡b, a = 4, b = 5 [b > a]  Dc‡Kw›`aK j‡¤^i •`N© ̈ = 2a2 b = 2  16 5 = 32 5 GKK 15. x 2 2 + y 2 2 = 2 Dce„‡Ëi Dc‡Kw›`aK j‡¤^i •`N ̈© KZ GKK? [What is the length of the latus rectum of the ellipse x 2 2 + y 2 2 = 2?] 4 3 2 2 2 4 3 DËi: 2 2 e ̈vL ̈v: Dce„‡Ëi mgxKiY, x 2 2 + y 2 2 = 2  x 2 2 2 + y 2 4 = 1; GLv‡b, a = 2 2, b = 2  b > a  Dc‡Kw›`aK j‡¤^i •`N© ̈ = 2a2 b = 2  2 2 2 = 2 2 GKK 16. xy = 2 mgxKiYwU n‡eÑ [The equation xy = 2 represents a:] cive„Ë (parabola) e„Ë (circle) Dce„Ë (ellipse) Awae„Ë (hyperbola) DËi: Awae„Ë e ̈vL ̈v: mgxKiYwU xy = 2; hv‡K ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 KwY‡Ki Av`k© mgxKi‡Yi mv‡_ Zzjbv K‡i cvB, 2h = 1  h = 1 2 a = 0 b = 0  ab – h 2 = 0 – 1 4 = – 1 4 < 0  xy = 2 Awae„‡Ëi mgxKiY| 17. 9x2 – 16y2 – 36x – 32y – 124 = 0 mgxKiY m~wPZ eμ‡iLvwU Kx wb‡`©k K‡i? [The equation 9x2 - 16y2 - 36x - 32y - 124 = 0 represents which curve?] Dce„Ë (ellipse) Awae„Ë (hyperbola) cive„Ë (parabola) None DËi: Awae„Ë (hyperbola) e ̈vL ̈v: A =       a h g h b f g f c =       9 0 – 18 0 – 16 – 16 – 18 – 16 – 124 a = 9, b = – 16, g = – 18, h = 0, f = – 16, c = – 124 = 9{(– 16)  (– 124) – 16  16} – 0 – 18(0 – 16  18)  0 ab = – 144 h = 0  ab – h 2 = – 144 – 0 = – 144 < 0  mgxKiYwU Awae„Ë n‡e|  = 0 n‡j hyMj mij‡iLv n‡Zv| 18. Avo A‡ÿi •`N ̈© 8 GKK Ges ( 12, 0) Dc‡K›`aØq wewkó Awae„‡Ëi Dr‡Kw›`aKZv n‡”QÑ [If the length of the transverse axis of a hyperbola is 8 units and the foci are (±12, 0), what is the eccentricity of the hyperbola?] 1 3 3 2 1 2 DËi: 3 e ̈vL ̈v: hw` a > b nq
4 Avo A‡ÿi •`N© ̈, 2a = 8  a = 4 Dc‡K›`aØq ( ae, 0)  ( 12, 0)  ae = 12  e = 12 4 = 3 19. †h cive„‡Ëi Dc‡K‡›`ai ̄’vbv1⁄4 (4, 0) Ges wbqvgK †iLvi mgxKiY, x + 2 = 0 Zvi mgxKiYÑ [If the focus of a parabola is (4, 0) and the equation of its directrix is x + 2 = 0, what is its equation?] y 2 = 6(x – 2) y 2 = 10(x – 3) y 2 = 12(x – 1) y 2 = 4(x – 1) DËi: y 2 = 12(x – 1) e ̈vL ̈v: awi, cive„‡Ëi Dci ̄’ †h‡Kv‡bv we›`y P(x, y) Ges Dc‡K›`a S(4, 0)  cive„‡Ëi mgxKiY, SP = PM  (x – 4) 2 + (y – 0) 2 =     x + 2 1 2 + 02  (x – 4)2 + y2 = (x + 2)2  x 2 – 8x + 16 + y2 = x 2 + 4x + 4  y 2 – 12x + 12 = 0  y 2 = 12x – 12 = 12(x – 1) 20. 9x2 – 16y2 – 18x – 64y – 199 = 0 Awae„‡Ëi AmxgZU؇qi †Q`we›`yi ̄’vbv1⁄4 KZ? [What are the coordinates of the point of intersection of the asymptotes of the hyperbola 9x2 - 16y2 - 18x - 64y - 199 = 0?] (1, – 2) (2, 1) (1, 2) (– 2, 1) DËi: (1, – 2) e ̈vL ̈v: AmxgZU؇qi †Q`we›`y = Awae„‡Ëi †K›`a| GLb, 9x2 – 16y2 – 18x – 64y – 199 = 0  9(x2 – 2x + 1) – 16(y2 + 4y + 4) = 199 + 9 – 64  9(x – 1)2 – 16(y + 2)2 = 144  (x – 1) 2 16 – (y + 2) 2 9 = 1  X 2 4 2 – Y 2 3 2 = 1  Awae„‡Ëi †K›`a (X, Y) = (0, 0)  x – 1 = 0  x = 1 Avevi, y + 2 = 0  y = – 2  †K›`a (1, – 2) 21. y 2 9 – x 2 4 = 1 Awae„‡Ëi AmxgZU؇qi mgxKiY †KvbwU? [What is the equation of the asymptotes of the hyperbola y 2 9 – x 2 4 = 1?] y =  2 3 x x =  3 2 y x =  2 3 y y =  3x DËi: x =  2 3 y e ̈vL ̈v: Awae„Ë, (y – 0) 2 3 2 – (x – 0) 2 2 2 = 1 AmxgZ‡Ui mgxKiY, x –  =  a b (y – )  x – 0 =  2 3 (y – 0)  3x =  2y  y =  3 2 x  x =  2 3 y 22. 5r2 sin2  = 12rcosn‡j, Dc‡K‡›`ai ̄’vbv1⁄4 wb‡Pi †KvbwU? [If 5r2 sin2θ = 12rcosθ, what are the coordinates of the epicenter?]     5 3  0 (0, 0)     3 5  0     0  3 5 DËi:     3 5  0 e ̈vL ̈v: 5r2 sin2  = 12rcos  5y2 = 12x  y 2 = 12 5 x  y 2 = 4 . 12 5  4 x  y 2 = 4 3 5 x     3 5  0 Shortcut: 5y2 = 12x  y 2 = 4 3 5 x     3 5  0  Dc‡K‡›`ai ̄’vbv1⁄4 (a, 0) =     3 5  0 23. x = 3cos + 2, y = sin + 1 Dce„‡Ëi †ÿÎdj KZ eM© GKK? [What is the area of the ellipse represented by the parametric equations x = 3cosθ + 2, y = sinθ + 1?] 3  3 3  DËi: 3 e ̈vL ̈v: x – 2 = 3cos x – 2 3 = cos ..........(i) y – 1 = sin ............(ii) (i)2 + (ii)2  (x – 2) 2 9 + (y – 1)2 = 1 †hvMdj = ab = 3  1 = 3