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

1 Varsity Daily-10 [Set-B (Solve Sheet)] wm‡jevm : KwbK c~Y©gvb: 30 †b‡MwUf gvK©: 0.25 mgq: 20 wgwbU 1. 2y = 3kx + 4 mij‡iLvwU y 2 = 32x cive„ˇK ̄úk© Ki‡j, k Gi gvb KZ? [If the line 2y = 3kx + 4 is tangent to the parabola y2 = 32x, what is the value of k?] 16 3 – 4 3 – 8 3 8 3 DËi: 8 3 e ̈vL ̈v: mij‡iLv 2y = 3kx + 4  y = 3 2 kx + 2 cive„Ë, y 2 = 32x  y 2 = 4  8  x; †hLv‡b a = 8 Avgiv Rvwb, y = mx + c †iLvwU y 2 = 4ax cive„‡Ëi ̄úk©K n‡j, c = a m  2 = 8 3k 2  2  3k 2 = 8  k = 8 3 2. y 2 = 8x cive„‡Ëi Dc‡K‡›`ai ̄’vbv1⁄4 †KvbwU? [What are the coordinates of the focus of the parabola y2 = 8x?] (0, 2) (– 2, 0) (0, – 2) (2, 0) DËi: (2, 0) e ̈vL ̈v: y 2 = 8x  y 2 = 4  2  x; †hLv‡b, a = 2 y 2 = 4ax cive„‡Ëi Dc‡K‡›`ai ̄’vbv1⁄4 (a, 0)  y 2 = 8x cive„‡Ëi Dc‡K‡›`ai ̄’vbv1⁄4 (2, 0) 3. 9x2 + 5y2 = 45 Dce„‡Ëi kxl©Ø‡qi ga ̈eZ©x `~iZ¡ KZ GKK? [What is the distance between the vertices of the ellipse 9x2 + 5y2 = 45?] 2 5 9 6 5 DËi: 6 e ̈vL ̈v: Dce„Ë, 9x2 + 5y2 = 45  x 2 5 + y 2 9 = 1  x 2 ( 5) 2 + y 2 3 2 = 1 [†hLv‡b a = 5 Ges b = 3; b > a] kxl©Ø‡qi ga ̈eZx© `~iZ¡ = 2b = 2  3 = 6 GKK a > b n‡j, kxl©Ø‡qi ga ̈eZx© `~iZ¡ 2a 4. 16x2 + 9y2 = 144 Dce„‡Ëi Dc‡Kw›`aK j‡¤^i •`N© ̈ KZ? [What is the length of the latus rectum of the ellipse 16x2 + 9y2 = 144?] 9 4 GKK ( 9 4 unit) 32 3 GKK ( 32 3 unit) 9 2 GKK ( 9 2 unit) 9 8 GKK ( 9 8 unit) DËi: 9 2 GKK ( 9 2 unit) e ̈vL ̈v: 16x2 + 9y2 = 144  x 2 9 + y 2 16 = 1  x 2 3 2 + y 2 4 2 = 1 †hLv‡b a = 3, b = 4 [b > a]  Dc‡Kw›`aK j‡¤^i •`N© ̈ = 2a2 b = 2  9 4 = 18 4 = 9 2 GKK| 5. y = 3x + 1 †iLvwU y 2 = 4ax cive„ˇK ̄úk© Ki‡j, Zvi †dvKvm KZ n‡e? [If the line y = 3x + 1 is tangent to the parabola y2 = 4ax, what is the focus of the parabola?] (0, 3) (3, 0) (– 3, 0) (0, – 3) DËi: (3, 0) e ̈vL ̈v: y = 3x + 1 †hLv‡b, m = 3 Ges c = 1 y = mx + c †iLvwU y 2 = 4ax cive„ˇK ̄úk© Ki‡j, c = a m  a = cm = 3  1 = 3  Dc‡K›`a (a, 0) = (3, 0) 6. (x + 4) 2 100 + (y – 2) 2 64 = 1 Dce„‡Ëi Dr‡Kw›`aKZv KZ? [What is the eccentricity of the ellipse (x + 4) 2 100 + (y – 2) 2 64 1?] 4 5 5 3 1 3 5 DËi: 3 5 e ̈vL ̈v: (x + 4) 2 100 + (y – 2) 2 64 = 1 Dce„‡Ëi a = 10 Ges b = 8; †hLv‡b a > b  Dr‡Kw›`aKZv, e = 1 – b 2 a 2 = 1 – 64 100
2 = 36 100 = 6 10 = 3 5 7. †Kv‡bv Kwb‡Ki Dr‡Kw›`aKZv 0 < e < 1 n‡j, H Kwb‡Ki Av`k© mgxKiY †KvbwU? [If the eccentricity of a conic section is 0 < e < 1, what is the standard equation of that conic?] x 2 a 2 – y 2 b 2 = 1 x 2 a 2 + y 2 b 2 = 1 y 2 b 2 – x 2 a 2 = 1 y 2 = 4ax DËi: x 2 a 2 + y 2 b 2 = 1 e ̈vL ̈v: Avgiv Rvwb, Dce„‡Ëi Dr‡Kw›`aKZ 0 < e < 1 Ges Dce„‡Ëi mgxKiY, x 2 a 2 + y 2 b 2 = 1 †hLv‡b, a > b A_ev b > a 8. 15x2 + 15y2 – 3xy + 24y – 265 = 0 eμ †iLvwUi R ̈vwgwZK cwiPq wK? [What is the geometric identity of the curve 15x2 + 15y2 - 3xy + 24y - 265 = 0?] cive„Ë (parabola) Dce„Ë (ellipse) e„Ë (circle) Awae„Ë (hyperbola) DËi: Dce„Ë (ellipse) e ̈vL ̈v: mgxKi‡Y, a = 15, b = 15, h = – 3 2  ab – h 2 = 15  15 –    –  3 2 2 = 225 – 9 4 > 0 †h‡nZz ab – h 2 > 0. myZivs mgxKiYwU GKwU Dce„Ë| 9. y 2 = 9x cive„‡Ëi Dci Aew ̄’Z GKwU we›`y P Gi †KvwU 12, P Gi Dc‡Kw›`aK `~iZ¡ KZ? [If the ordinate of a point P on the parabola y2 = 9x is 12, what is the point's eccentric distance?] 19 1 2 16 18 1 4 10 DËi: 18 1 4 e ̈vL ̈v: y 2 = 9x  (12)2 = 9x [†h‡nZz †KvwU 12]  9x = 144  x = 16 y 2 = 4ax cive„‡Ëi Dc‡Kw›`aK `~iZ¡ x + a y 2 = 9x = 4  9 4  x  a = 9 4  x + 9 4 = 16 + 9 4 = 64 + 9 4 = 73 4 = 18 1 4 10. x 2 2 + y 2 3 = 1 Dce„Ëxq †ÿ‡Îi †h Ask abvZ¥K e„nr Aÿ I ÿz`a Aÿ Øviv †ewóZ Zvi †ÿÎdj KZ eM© GKK? [What is the area of the region enclosed by the ellipse x 2 2 + y 2 3 = 1 and its major axis?]  3 2  6 6 4 3 2  DËi: 3 2  e ̈vL ̈v: Dce„Ë, x 2 2 + y 2 3 = 1  x 2 ( 2) 2 + y 2 ( 3) 2 = 1  m¤ú~Y© Dce„‡Ëi †ÿÎdj =   2  3 = 6eM© GKK abvZ¥K e„nr Aÿ I ÿz`a Aÿ Øviv †ewóZ As‡ki †ÿÎdj =  6 2 =   3  2 2  2 =   3 2 = 3 2  eM© GKK| 11. x 2 – y 2 = 18 Awae„‡Ëi †dvKvm؇qi ga ̈eZx© `~iZ¡ KZ? [What is the distance between the foci of the hyperbola x2 - y2 = 18?] 2 2 GKK (2 2 unit) 3 GKK (3 unit) 12 GKK (12 unit) 2 6 GKK (2 6 unit) DËi: 12 GKK (12 unit) e ̈vL ̈v: x 2 – y 2 = 18  x 2 18 – y 2 18 = 1  x 2 (3 2) 2 – y 2 (3 2) 2 = 1 †dvKvm؇qi ga ̈eZx© `~iZ¡ 2ae = 2  a 2 + b2 = 2  18 + 18 = 2  36 = 2  6 = 12 GKK 12. Awae„‡Ëi Dc‡K›`a؇qi ga ̈eZx© `~iZ¡ 12 GKK Ges e = 2 n‡j, mgxKiY †KvbwU? [If the distance between the foci of a hyperbola is 12 units and its eccentricity is 2, what is its equation?] y 2 – x 2 = 18 x 2 – y 2 = 18 2x2 – y 2 = 4 x 2 4 – y 2 2 = 1 DËi: x 2 – y 2 = 18
3 e ̈vL ̈v: awi, Awae„Ë x 2 a 2 – y 2 b 2 = 1 Dc‡K›`a؇qi `~iZ¡ 2ae = 12  ae = 6  a = 6 2  a = 2  2  3 2  a = 3 2  a 2 = 18 Avevi, e = 1 + b 2 a 2  1 + b 2 18 = 2  1 + b 2 18 = 2  b 2 18 = 1  b 2 = 18 Awae„‡Ëi mgxKiY: x 2 18 – y 2 18 = 1  x 2 – y 2 = 18 13. y 2 = – x Gi w`Kv‡ÿi mgxKiY †KvbwU? [What is the equation of the directrix of the parabola y2 = -x?] 4x + 1 = 0 4x – 1 = 0 4y + 1 = 0 4y – 1 = 0 DËi: 4x – 1 = 0 e ̈vL ̈v: y 2 = 4     –  1 4 x a = – 1 4 y 2 = 4ax cive„‡Ëi w`Kv‡ÿi mgxKiY x = – a  x = –    –  1 4  x = 1 4  4x = 1  4x – 1 = 0 14. y 2 = 12ax cive„ËwU (3, – 2) we›`yMvgx n‡j, cive„ËwUi Dc‡Kw›`aK j‡¤^i •`N© ̈ KZ GKK? [If the parabola y2 = 12ax passes through the point (3, 2), what is the length of its latus rectum?] 9 4 4 3 4 9 2 3 DËi: 4 3 e ̈vL ̈v: y 2 = 12ax cive„ËwU (3, – 2) we›`yMvgx|  (– 2)2 = 12a  3  36a = 4  a = 1 9  y 2 = 12  1 9  x = 4  1 3  x  Dc‡Kw›`aK j‡¤^i •`N ̈© = |4a| = 4  1 3 = 4 3 GKK 15. x 2 + 3y = 0 mgxKi‡Yi †jLwPÎ wb‡Pi †KvbwU? [Which of the following graphs represents the equation x2 + 3y = 0?] x x y y O x x y y O x x y y O x x y y O DËi: x x y y O e ̈vL ̈v: x 2 + 3y = 0  x 2 = – 3y cive„‡Ëi kxl© (0, 0)| †jLwPÎ x A‡ÿi FYvZ¥K w`‡K n‡e A_©vr x 2 = – 4ay AvKv‡ii n‡e| 16. 3y2 – x 2 = 9 Awae„‡Ëi Dr‡Kw›`aKZv KZ? [What is the eccentricity of the hyperbola 3y2 - x2 = 9?] 1 2 2 2 3 3 2 3 DËi: 2 e ̈vL ̈v: Awae„‡Ëi mgxKiY, 3y2 – x 2 = 9  y 2 3 – x 2 9 = 1  y 2 ( 3) 2 – x 2 (3) 2 = 1; a = 3, b = 3  Dr‡Kw›`aKZv, e = 1 + a 2 b 2 = 1 + 9 3 = 12 3 = 4 = 2 17. hw` y 2 = 18x †Kvb cive„Ë nq, Zvn‡j P(2, 4) we›`yi †dvKvm `~iZ¡ KZ GKK n‡e? [If y2 = 18x is the equation of a parabola and P(2, 4) is a point on the parabola, what is the focal distance of point P?] 4 26 13 2
4 25 4 6 DËi: 13 2 e ̈vL ̈v: cive„Ë, y 2 = 18x = 4  9 2  x     a = 9 2 y 2 = 4ax cive„‡Ëi †dvKvm `~iZ¡ = x + a P(2, 4) we›`yi Rb ̈ †dvKvm `~iZ¡ = 2 + a = 2 + 9 2 = 13 2 GKK 18. 25x2 + 16y2 = 400 Dce„‡Ëi w`Kv‡ÿi mgxKiY †KvbwU? [What is the equation of the directrix of the ellipse 25x2 + 16y2 = 400?] y =  5 3 x =  20 3 y =  25 3 x =  32 5 DËi: y =  25 3 e ̈vL ̈v: Dce„‡Ëi mgxKiY, 25x2 + 16y2 = 400  x 2 16 + y 2 25 = 1  x 2 4 2 + y 2 5 2 = 1 [b > a]  w`Kv‡ÿi mgxKiY, y =  b e Dr‡Kw›`aKZv, e = 1 – a 2 b 2 [⸪ b > a] = 1 – 16 25 = 9 25 = 3 5  w`Kv‡ÿi mgxKiY, y =  5 3 5 =  25 3 19. 5x2 + 15x – 10y – 4 = 0 cive„‡Ëi wbqvg‡Ki mgxKiY †KvbwU? [What is the equation of the directrix of the parabola 5x2 + 15x – 10y – 4 = 0?] 40y – 81 = 0 40x + 81 = 0 40y + 81 = 0 40y + 41 = 0 DËi: 40y + 81 = 0 e ̈vL ̈v: 5x2 + 15x – 10y – 4 = 0  x 2 + 3x – 2y – 4 5 = 0  x 2 + 3x = 2y + 4 5  x 2 + 2 . 3 2 x +     3 2 2 = 2y + 4 5 + 9 4      x + 3 2 2 = 2y + 16 + 45 20 = 2y + 61 20 = 2    y + 61 40     x + 3 2 2  1 2    y + 61 40 [a = 1 2 ] x 2 = 4ay cive„‡Ëi wbqvg‡Ki mgxKiY: y = – a y + 61 40 = – 1 2  y = – 1 2 – 61 40  40y + 81 = 0 20. y 2 – 6y + 4x + 25 = 0 GKwU cive„‡Ëi mgxKiY n‡j, w`Kv‡ÿi cv`we›`yi ̄’vbv1⁄4 KZ n‡e? [If y2 - 6y + 4x + 25 = 0 is the equation of a parabola, what are the coordinates of the foot of its directrix?] (3, – 3) (3, 3) (– 3, 3) (– 5, 3) DËi: (– 3, 3) e ̈vL ̈v: y 2 – 6y + 4x + 25 = 0  y 2 – 6y = – 4x – 25  y 2 – 2y.3 + 32 = – 4x – 25 + 32  (y – 3)2 = – 4x – 16  (y – 3)2 = – 4(x + 4) = 4  (– 1)  (x + 4) w`Kv‡ÿi cv`we›`yi ̄’vbv1⁄4 (– a, 0) = (x + 4, y – 3)  x + 4 = + 1  x = – 3 y – 3 = 0  y = 3  ̄’vbv1⁄4 (– 3, 3) 21. 3r2 cos2  – 4rsin + 6rcos–Dc‡Kw›`aK j‡¤^i •`N© ̈ KZ? [If 3r2cos2θ - 4rsinθ + 6rcosθ - 5 = 0 is the equation of a conic section, what is the length of its latus rectum?] – 3 4 3 4 3 + x 4 4 3 DËi: 4 3 e ̈vL ̈v: 3x2 – 4y + 6x – 5 = 0  3x2 + 6x = 4y + 5  x 2 + 2x = 4y + 5 3  (x + 1)2 = 4 . 1 3 (y + 2)  Dc‡Kw›`aK j¤^ = 4a = 4 1 3 = 4 3 22.     4  2 3 Ges (3, 2) we›`y GKwU Dce„‡Ëi AÿØq, ̄’vbv‡1⁄4i AÿØq eivei n‡j, Dce„ËwUi Dr‡Kw›`aKZv KZ?