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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
`y^(2)=-8x` |
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Answer» Given equation `y^(2)=-8x` Comparing with `y^(2)=-4ax` `4a=8" "rArr" "a=2` `:. "Coordinates of focus"-=(-a,0)-=(-2,0)` Axis of parabola of directrix : `x-a=0" "rArr" "a-2=0` Length of latus rectum `=4a=4xx2=8` |
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| 52. |
`x^(2)=-16y` |
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Answer» Given equation `x^(2)=-16y` Comparing with `x^(2)=-4ay`, `4a=16" "rArr" "a=4` `:. "Coordinates of focus"-=(0,-a)-=(0,-4)` Axis of parabola : x=0 Equation of directrix : y-a=0 `rArr" "y-4=0` Length of latus rectum `=4a=4xx4xx=16` |
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| 53. |
Focus (0,-3), directrix y=3 |
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Answer» Focus `S-=(0,-3)` and P(x,y) be a veriable point on parabola. `:.` Now, distance of P from focus S = perpendicular distance from P to the directrix `rArr" "sqrt((x-0)^(2)+(y+3)^(2)-y-3)` `rArr" "x^(2)+(y+3)^(2)=(y-3)^(2)` `rArr=-12y` |
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| 54. |
`y^(2)=10x` |
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Answer» Given equation `y^(2)=10x` Comparing with `y^(2)=4ax` `4a=10" "rArr" "a=(5)/(2)` `:. "Coordinates of focus"=(a,0)=((5)/(2),0)` Axis of parabola : y=0 Equation of directrix : x+a=0 `rArr""x+(5)/(2)=0" "rArr2x+5=0` Length of latus rectum `=4a=4xx(5)/(2)=10` |
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| 55. |
Focus (6,0), directrix x=-6 |
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Answer» Focus `S-=(6,0)` and directrix `x=-6" "rArr" "x+6=0` Let P(x,y) be a variable point on the parabola. `:.` Distance of from focus S = Perpendicular distance from P to the directrix. `rArrsqrt((x-6)^(2)+(y-0)^(2)=x+6)` `rArr" "(x-6)^(2)+y^(2)=(x+6)^(2)` `rArr" "y^(2)=24x` |
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| 56. |
The eccentricity of the curve `x^(2)-4x+4y^(2)=12` isA. `sqrt(3)`B. `(1)/(sqrt(3))`C. `(2)/(3)`D. None of these. |
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Answer» Correct Answer - D |
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| 57. |
The co-ordinates of a focus of an ellipse is (4,0) and its eccentricity is `(4)/(5)` Its equation is :A. `25x^(2)+9y^(2)=225`B. `9x^(2)+25y^(2)=225`C. `16x^(2)+25y^(2)=400`D. `25x^(2)+16y^(2)=400`. |
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Answer» Correct Answer - B |
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| 58. |
`x^(2)=-9y` |
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Answer» Given equation `x^(2)=-9y` Comparing with `x^(2)=-4ay`, `4a=9" "rArr" "a=(9)/(4)` Coordinates of focus `-=(0,-a)-=(0,-(9)/(4))` Axis of parabola : x=0 Equation of directrix : y-a=0 `rArr" "y-(9)/(4)=0` `rArr" "4y-9=0` Length of latus rectum =4a=9 |
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| 59. |
Find the centre and radius of the circle `3x^(2) 3y^(2) - 6x +9y - 8 =0`. |
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Answer» `3x^(2)+3y^(2)-69y-8=0` `rArrx^(2)+y^(2)-2x+3y-(8)/(3)=0` Comparing with the general equation of the circle `x^(2)+y^(2) +2gx +c =0` `2g=-2,2f=3 and c=-(8)/(3)` `rArrg=-1,f=(3)/(2)and c=-(8)/(3)` `:.` Centre of circle `= (-g,-f)` `= (1,-(3)/(2))` And radius of circle `=sqrt(g^(2)+f^(2)-c)=sqrt(1+(9)/(4)-(-(8)/(3)))` `=sqrt(1+(9)/(4)+(8)/(3))=sqrt((12+27+32)/(12))` `= sqrt((71)/(12))=(sqrt(71))/(2sqrt(3))` units. |
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| 60. |
Find the diameter of the circle `2x^(2)+2y^(2)-6x-9=0`. |
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Answer» Correct Answer - `3sqrt(3)` |
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| 61. |
Find the equation of a circle whose centre is (2, 3) and radius is 5. |
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Answer» Here centre `(h,k) = (2,3)`, and radius `a=5` `:.` Equation of circle `(x-h)^(2)+(y-k)^(2)=a^(2)` `rArr (x-2)^(2) + (y-3)^(2)=5^(2)` `rArr x^(2)-4x+4 y^(2)-6y +9 = 25` `rArr x^(2)+y^(2)-4x -6y - 12=0` |
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| 62. |
Find the centre and radius of the circle `(x-2)^(2)+(y+5)^(2)=49` |
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Answer» `(x-2)^(2)+(y+5)^(2)=49` `rArr(x-2)^(2)+(y-5)^(2)=7^(2)` Comparing with the equation of circle `(x-h)^(2)+(y-k)^(2)-a^(2)` `h=2,k=-5,a=7` `:.` Centre of circle `-(h,k)=(2,-5)` and radius `a=7`. |
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| 63. |
Find the centre and radius of the circle `(x+2)^(2)+(y+3)^(2)=5` |
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Answer» `(x+2)^(2)+(y+3)^(2)=5` `rArr(x+2)^(2)+(y+3)^(2)=(sqrt(5))^(2)` Comparing with the equation of the circle `(x-h)^(2)+(y-k)^(2)=a^(2)` `h=-2,k=-3,a=sqrt(5)` `:.` Centre of circle `=(h,k) =(-2,-3)`, and radius `a= sqrt(5)`. |
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| 64. |
Find the equation of the ellipse whose foci are `(pm3,0)` and it passes through the point `(2,sqrt(7))`. |
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Answer» Correct Answer - `x^(2)+2y^(2)=18` |
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| 65. |
Find the equation of the ellipse that satisfies the given conditions: Vertices `(pm5,0)`, foci `(pm4,0)` |
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Answer» Vertices `-=(pm5,0)`, foci `(pm4,0)` `because` The y-coordinate of each of vertices and foci are zero. `:.` The major axis of the ellipse will be along x-axis. Now, a=5 and ae=4 and `b^(2)=a^(2)(1-e^(2))` `=a^(2)-a^(2)e^(2)` `=5^(2)-4^(2)=9` Equation of ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 rArr(x^(2))/(25)+(y^(2))/(9)=1` |
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| 66. |
Find the equation of a circle, the co-ordinates of the ends of whose diameter are `(-1,-3) and (2,5)` Now the equation of circle `(x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0` `rArr(x+1)(x-2)+(y+3)(y-5)=0` `rArr x^(2)-x-2+ y^(2)-2y-15=0` `rArr x^(2)+y^(2)-x-2y-17=0` We is the required equation. |
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Answer» Here `(x_(1),y_(1))=(-1,-3) and (x_(2),y_(2))=(2,5)` Now the equation of circle `(x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0` `rArr(x+1)(x-2)+(y+3)(y-5)=0` `rArr x^(2)-x-2+ y^(2)-2y-15=0` `rArr x^(2)+y^(2)-x-2y-17=0` which is the required equation. |
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| 67. |
Find the equation of the ellipse that satisfies the given conditions:Foci `(pm3,0),a=4` |
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Answer» Here, `a=4rArra^(2)=16` Foci `-=(pm3,0)` are on x-axis `:.` The major axis will be along x-axis. Now, ae=3 and `b^(2)=a^(2)(1-e^(2))` `=a^(2)-a^(2)e^(2)=4^(2)-3^(2)=7` `:.` Equation of ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1rArr(x^(2))/(16)+(y^(2))/(7)=1` |
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| 68. |
Find the equation of the ellipse whose vertices are `(+-13 ,0)`and foci are `(+-5,0)`. |
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Answer» The vertices and foci of the ellipse are on x-axis, therefore let the equation of ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` where `agtb` Now given that, vertices `=(pma,0)=(pm13,0)` a=13 and `foci=(pmae,0)=(pm5,0)` `rArr" "ae=5` `rArr" "13e=5` `rArr" "e=(5)/(13)` Now `b^(2)=a^(2)(1-e^(2))` `=169(1-(25)/(169))=144` `:.` Equation of ellipse `(x^(2))/(169)+(y^(2))/(144)=1`. |
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| 69. |
Find the equation of the ellipse that satisfies the given conditions:Length of minor axis 16, foci `(0pm6)`. |
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Answer» Here, foci `-=(0,pm6)` are on y-axis. `:.` Major axis is along y-axis. Now be = 6 Given, minor axis 2b = 16 `rArr" "b=8" "rArr" "b^(2)=64` and `a^(2)=b^(2)(1-e^(2))` `rArr" "a^(2)=b^(2)-b^(2)e^(2)` `rArr" "a^(2)=64-(6)^(2)=28` Now, equation of ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1rArr(x^(2))/(28)+(y^(2))/(64)=1` |
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| 70. |
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse : `(x^(2))/(16)+(y^(2))/(9)=1` |
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Answer» `(x^(2))/(16)+(y^(2))/(9)=1` `:.` Comparing with `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` `:.a^(2)=16" "rArr" "a=4` `b^(2)=9" "rArr" "b=3` `:." "agtb`. The major aixs of the ellipse will be along x-axis. Vertices `-=(pma,0)-=(pm4,0)` Major axis = ab = 2(3) = 6 Eccentricity `e=sqrt(1-(b^(2))/(a^(2)))=sqrt(1-(1)/(16))=(sqrt(7))/(4)` coordinates of foci `-=(pmae,0)-=(pmsqrt(7),0)` Latus rectum `=(2b^(2))/(a)=(2xx9)/(4)=(9)/(2)` |
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| 71. |
Find th equation of the hyperbola eccentricity is 2 and the distance between foci 16. |
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Answer» Here e 2 Distance between foci =2ae=16 `rArr" "2a(2)=16` `rArr" "a=4` and `b^(2)=a^(2)(e^(2)-1)` `=16(4-1)=48` `:.` Equation of hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` `rArr" "(x^(2))/(16)-(y^(2))/(48)=1` `rArr" "3x^(2)-y^(2)=48`. |
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| 72. |
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse `(x^(2))/(4)+(y^(2))/(25)=1` |
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Answer» `(x^(2))/(4)+(y^(2))/(25)=1` Comparing with `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` `:.a^(2)=4andb^(2)=25" "rArr" "a=2andb=5` Here, `altb`. `:.` The major axis of the ellipse will be along y-axis Vertices `-=(0,pmb)-=(0,pm5)` Eccentricity `e=sqrt(1-(a^(2))/(b^(2)))` `=sqrt(1-(4)/(25))=sqrt((21)/(25))=(sqrt(21))/(5)` Coordinates of foci `-=(0,pmbe)` `-=(0,pmxx(sqrt(21))/(5))-=(0,pmsqrt(21))` Major axis `=2b=2xx5=10` Minor axis `=2a=2xx2=4` Length of latus rectum `=(2a^(2))/(b)=(2xx4)/(5)=(8)/(5)`. |
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| 73. |
Find the equation of the hyperbola where foci are `(0,+-12)`and the length of the latus rectum is 36. |
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Answer» The focus of the hyperbola lies on y-axis, terefore, let the equation of the hyperbola is `(y^(2))/(b^(2))-(x^(2))/(a^(2))=1` Co-ordinates of foci `=(0,pmbe)` be=12 and latus rectum `(2a^(2))/(b)=36` `rArr" "a^(2)=18b` `rArr" "b^(2)(e^(2)-1)=18b` `rArr" "144-b^(2)=18b` `rArr""b^(2)+18b-144=0` `rArr" "(b+24)(b-6)=0` `rArr" "b=-24orb=6` b=-24 (which is not possible) `:.` b=6 Therefore, equation of hyerbola `(y^(2))/(36)-(x^(2))/(108)=1` `rArr" "3y^(2)-x^(2)=108`. |
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| 74. |
`(x^(2))/(36)+(y^(2))/(16)=1` |
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Answer» Equation of ellipse : `(x^(2))/(36)+(y^(2))/(16)=1` Comparing with `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` `:." "a^(2)=36andb_(2)=16` `rArr" "a=6andb=4` `"Here",agtb`. `:.` Major axis of the ellipse will along x-axis. Vertices `-=(pma,0)-=(pm6,0)` Eccentricity `e=sqrt(1-(b^(2))/(a^(2)))=sqrt(1-(16)/(36))=sqrt((20)/(36))` `rArr" "e=(sqrt(5))/(3)` coordinates of foci `-=(pmae,0)` `-=(pm6xx(sqrt(5))/(3),0)-=(pm2sqrt(5,)0)` Major axis `=2a2xx6=12` Minor axis `=2b=2xx4=8` Length of latus rectum `=(2b^(2))/(a)=(2xx16)/(6)=(16)/(3)` |
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| 75. |
Vertex (0,0), passing through (5,2) and symmetric with respect to y-axis |
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Answer» Let vertex `-=(0,0)` and equation of parabola symmetric to y-axis be `x^(2)=4ay` It passes through the point (5,2). `:." "25=4a(2)` `rArr" "4a=(25)/(2)` Therefore, equation of parabola is `x^(2)=(25)/(2)y`. |
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| 76. |
Vertex (0, 0) passing through (2,3) and axis is along x-axis. |
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Answer» Vertex = (0,0) and axis is along x-axis. `:.` Let equation of parabola is `y^(2)=4ax`. Point (2,3) lies on the parabola. `:." "(3)^(2)=4a(2)` `rArr" "4a=(9)/(2)` Now, equation of parabola is `y^(2)=(9x)/(2)" "rArr" "2y^(2)=9x` |
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| 77. |
Find the equation of circumcircle of the triangle whose sides are `2x +y = 4,x + y = 6 and x +2y=5`. |
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Answer» First we will find the co-ordinates of the vertices of a triangle `:.` On solving two-two equations, we get the vertices as `(-2,8),(7,-1) and (1,2)` Let the equation of circle be `x^(2)+y^(2)+2gx+2fy+c=0`....(1) This circle passes through the point `(-2,8)` `4+64 -4g +16f + c=0` `rArr -4g +16f +c = -68` .....(2) This circle passes through the point `(7,-1)` `49 +1 +14g-2f +c=0` `rArr 14g - 2f +c = -50` ....(3) This circle passes through the point (1, 2) `1+4+2g+4f +c=0` `rArr2g+4f+c=-5`....(4) From eqs. (2),(3) and (4) `g=-(17)/(2),f=-(19)/(2),c=50` From eq. (1), equation of circle `x^(2)+y^(2)-17x-19y+50=0`. |
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| 78. |
Find the equation of ellipse whose vertices are `(0,pm13)` and foci `(0,pm5)`. |
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Answer» The vertices and foci of ellipse are on y-axis. `:." "bgta` Now co-ordinates of the vertices = `0,pmb)` and co-ordinates of foci `=(0,pmbe)` `:.` b=13 and be=5 `:." "e=(5)/(13)` and `a^(2)=b^(2)(1-e^(2))` `=169(1-(25)/(169))=144` `:.` Therefore, the equation of ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` `rArr" "(x^(2))/(144)+(y^(2))/(169)=1`. |
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| 79. |
Find the equation of the ellipse that satisfies the given conditions:Length of major axis 26, foci `(pm5,0)` |
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Answer» Here, foci `-=(pm5,0)`, which lie on x-axis. `:.` Given, major axis 2a=26 `rArr" "a=13" "a^(2)=169` and `b^(2)=a^(2)(1-e^(2))` `=a^(2)-a^(2)e^(2)` `=169-(5)^(2)=144` Now, equation of ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1rArr(x^(2))/(169)+(y^(2))/(144)=1` |
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| 80. |
Find the equation of the ellipse that satisfies the given conditions: Vertices `(0,pm13)`, foci `(0,pm5)` |
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Answer» The foci and vertices of ellipse are on y-axis. `:. bgta` Coordinates of the vertices of ellipse `=(0,pmb)` and coordinates of foci `=(0,pmbe)` shows that b=13 and be = 5. `:." "e=(5)/(13)` `and" "a^(2)=b^(2)(1-e^(2))` `=169(1-(25)/(169))=144` `:.` Equation of ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1rArr(x^(2))/(144)+(y^(2))/(169)=1` |
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| 81. |
(i) Find the point at which the circle `x^(2)+y^(2)-5x+2y+6=0`, meets the X-axis. |
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Answer» Correct Answer - (2,0) , (3,0) |
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| 82. |
If `t` is the parameter for one end of a focal chord of the parabola `y^2 =4ax,` then its length is :A. `a(t+(1)/(t))`B. `a(t-(1)/(t))`C. `a(t+(1)/(t))^(2)`D. `a(t-(1)/(t))^(2)` |
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Answer» Correct Answer - C |
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| 83. |
`x^(2)=6y` |
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Answer» Given equation : `x^(2)=6y` Comparing with `x^(2)=4ay` `4a=6rArra=(3)/(2)` `:. "Coordinates of focus" -=(0,a)-=(0,(3)/(2))` Axis of parabola : : x=0 Equation of directrix :y+a=0 `rArry+(3)/(2)=0" "rArr" "2y+3=0` Length of latus rectum, 4a `=4xx(3)/(2)=6` |
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| 84. |
Prove that the circle `x^(2)+y^(2)+2x+2y+1=0` and circle `x^(2)+y^(2)-4x-6y-3=0` touch each other. |
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Answer» For first circle `x^(2)+y^(2)+2x+2y+1=0` `2g_(1)=2,2f_(1)=2,c_(1)=1` `rArrg_(1)=1,f_(1)=1,c_(1)=1` `:.` Centre `= P_(1)(-g_(1),-f_(1))=P_(1)(-1,-1)` and radius `=r_(1)=sqrt(g_(1)^(2)+f_(1)^(2)-c_(1))` `=sqrt(1+1-1)=1` units For second circle `x^(2)+y^(2)-4x-6y-3=0` `2g_(2)=-4,2f_(2)=-6,c_(2)=-3` `rArrg_(2)=-2,f_(2)=-3,c_(2)=-3` `:.` Centre `= P_(2)(-g_(2),-f_(2))=P_(2)(2,3)` and radius `r_(2)=sqrt(g_(2)^(2)+f_(2)^(2)-c_(2))` `=sqrt(4+9+3)=4` units Now the distance between the centre of circles `P_(1)P_(2)=sqrt((2+1)^(2)+(3+1)^(2))` `=sqrt(9+16)=sqrt(25)=5` `=r_(1)+r_(2)` Therefore, two circles touch each other. |
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| 85. |
(i) Find the equation of a circle concentric with the circle `x^(2)+y^(2)-8x+6y-10=0` and passes through the point (-2,3). (ii) Find the equation of circle concentric with the circle `x^(2)+y^(2)-4x-8x-6=0` and whose radius is three times the radius of this circle. |
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Answer» Correct Answer - (i) `x^(2)+y^(2)-8x+6y-47=0` , (ii) `x^(2)+y^(2)-4x-8y-214=0` |
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| 86. |
Find the equation of the hyperbola whose foci are `(0,pm4) and latus rectum is 12. |
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Answer» Correct Answer - `3y^(2)-x^(2)=12` |
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| 87. |
Find the equation of Hyperbola satisfying the following conditions:Foci `(pm3sqrt(5),0)` the latus rectum is of length 8. |
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Answer» Here, `ae=3sqrt(5)` and `(2b^(2))/(a)=8` `rArrb^(2)=4a` `rArra^(2)(e^(2)-1)=4a` `rArra^(2)e^(2)-1a^(2)=4a` `rArr45-a^(2)=4a` `rArra^(2)+4a-45=0` `rArr(a+9)(a-5)=0` `rArra=-9anda=5` But `a!=-9` `:." "a=5` `and" "b^(2)=4a=4xx5=20` Therefore, equation of hyperbola `(x^(2))/(25)-(y^(2))/(20)=1rArr4x^(2)-5y^(2)=100` |
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| 88. |
Find the equation of Hyperbola satisfying the following conditions: Foci `(pm4,0)`, the latus rectum is of length 12. |
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Answer» Foci `(pm4,0)`, lie on x-axis. `:." "ae=4rArra^(2)e^(2)=16` . . . (1) and latus rectum `=(2b^(2))/(a)=12` `rArr" "b^(2)=6a` . . .(2) `rArra^(2)(e^(2)-1)=6a` `rArra^(2)e^(2)-a^(2)=6a` `rArr" "16-a^(2)=6a` [From equation (1)] `rArr" "a^(2)+6a-16=0` `rArr" "(a+8)(a-2)=0` `rArr" "a=-8ora=2` But a cannot be negative. `:.a=2rArra^(2)=4` From equation (2) `b^(2)=6xx2=12` Now, equation of hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1rArr(x^(2))/(4)-(y^(2))/(12)=1` |
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| 89. |
Find the equations of the hyperbola satisfying the given conditions :Foci `(0,+-13)`, the conjugate axis is of length 24. |
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Answer» Here, foci `(0,pm13)` lie on y-axis `:." "be=13rArrb^(2)e^(2)=169` and conjugate axis 2a=24 `:." "a=12rArra^(2)=144` Now, `a^(2)=b^(2)(e^(2)-1)` `rArr" "a^(2)=b^(2)e^(2)-b^(2)` `rArr" "144=169-b^(2)` `rArr" "b^(2)=25` `:.` Equation of hyperbola `(y^(2))/(b^(2))-(x^(2))/(a^(2))=1rArr(y^(2))/(25)-(x^(2))/(144)=1` |
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| 90. |
Find the equation of Hyperbola satisfying the following conditions: Foci `(0,pmsqrt(10))`, passing through (2,3). |
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Answer» Let equation of hyperbola is `(y^(2))/(b^(2))-(x^(2))/(a^(2))=1` . . . (1) `(because` foci lie on y-axis) Coordinates of foci `=(0,pmsqrt(10))` `rArr" "be=sqrt(10)` `rArr" "b^(2)e^(2)=10` `rArra^(2)+b^(2)=10` . . .(2) Hyperbola passes through the point (2,3). Therefore from equation (1) `(9)/(b^(2))-(4)/(a^(2))=1` `(9)/(b^(2))-(4)/(10-b^(2))=1` [From equation (2)] `rArr(90-9b^(2)-ab^(2))/(b^(2)(10-b^(2)))` `rArr""90-13b^(2)=10b^(2)-b^(4)` `rArr""b^(4)-23b^(2)+90=0` `(b^(2)-5)(b^(2)-18)=0` `rArr""b^(2)=5orb^(2)=18` From equation (2) `b^(2)=5rArra^(2)=5` `b^(2)=18rArra^(2)=-8` which is not possible `:." "b^(2)=a^(2)=5` and equation of hyperbola `(y^(2))/(5)-(x^(2))/(5)=1rArry^(2)-x^(2)=5` |
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| 91. |
Find the equation of the ellipse whose foci are `(0,pm1)` and eccentricity is `(1)/(2)`. |
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Answer» Correct Answer - `(x^(2))/(3)+(y^(2))/(4)=1` |
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