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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.
| 1. |
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))/(100)+(y^(2))/(400)=1` |
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Answer» `(x^(2))/(100)+(y^(2))/(400)=1` Comparing with `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` `:.a^(2)=100" "rArr" "a=10` `b^(2)=400" "rArr" "b=20` Here, `altb`. `:.` the major axis of the ellipse wil be along y-axis. Vertices `-=(0,pmb)-=(0,pm20)` Eccentricity `e=sqrt(1-(a^(2))/(b^(2)))=sqrt(1-(100)/(400))=sqrt((3)/(4))=(sqrt(3))/(2)` Now, be `=20xx(sqrt(3))/(2)xx10sqrt(3)` `:. "Coordinates of foci" -=(0,pmbe)-=(0,pm10sqrt(3))` Major axis `=2b=2xx20=40` Minor axis `=2a=2xx10=20` Length of latus rectum `=(2a^(2))/(b)=(2xx100)/(20)=10` |
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| 2. |
If the focus of a parabola is `(3,3)` and its directrix is `3x-4y=2` then the length of its latus rectum isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B |
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| 3. |
Find the vertex, focus, directrix, latus rectum, equation of latus rectum, equation of axis and co-ordinates of ends of latus rectum for the following parabola : (i) `y^(2)=20x` , (ii)`y^(2)=-8y` (iii) `x^(2)=16y` , (iv) `x^(2)=-8y` (v) `2x^(2)=3y` , (iv) `3y^(2)+4x=0` |
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Answer» Correct Answer - (i) Vertex (0,0), focus (5,0), directrix x=-5, latus rectum = 20, equation of latus rectum x=5, equation of axis y=0, co-ordinates of the ends of latus rectum (5,10),(5-10). (ii) Vertex (0,0), focus (-3,0), directrix x=3, latus rectum = 12, equation of latus rectum x=-3, equation of axis y=0, co-ordinates of the ends of latus rectum (-3,6),(-3-6). (iii) Vertex (0,0), focus (0,4), directrix x=-4, latus rectum y = 16, equation of latus rectum y=4, equation of axis x=0, co-ordinates of the ends of latus rectum (8,4),(-8-4). (iv) Vertex (0,0), focus (0,-2), directrix x=2, latus rectum = 8, equation of latus rectum y=-2, equation of axis x=0, co-ordinates of the ends of latus rectum (4,-2),(-4,-2). (v) Vertex (0,0), focus `(0,(3)/(8))`, directri x `y=-(3)/(8)`, latus rectum `=(3)/(2)`, equation of latus rectum `y=(3)/(8)`, equation of axis x=0, co-ordinates of the ends of latus rectum `((-3)/(4),(3)/(8)),((3)/(4),(3)/(8))`. (vi) Vertex (0,0) focus `(-(1)/(3),0)`, directrix `x-(1)/(3)`, latus rectum `=(4)/(3)`, equation of latus rectum `x=-(1)/(3)`, equation of directrix y = 0, co-ordinates of the ends of latus rectum `(-(1)/(3),(2)/(3)),(-(1)/(3),-(2)/(3))`. |
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| 4. |
Find the vertex, focus, equation of axis, equation of axis, equation of directrix, length of latus rectum and the co-ordinates of the ends of latus rectum for the following parabolas : (i) `y^(2) =8x` (ii) `y^(2) = -12x` (iii) `x^(2)=6y` (iv) `x^(2)=-2y`. |
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Answer» (i) `y^(2) = 8x` Comparing with `y^(2) =4ax` `4a = 8 rArr a=2` `:.` Co-ordinates of vertex = (0, 0) Co-ordinates of focus `= (alpha,0)=(2,0)` Equation of axis `y=0` Equation of directrix `x = -a` `rArr x=-2` Length of latus rectum `= 4 alpha =8` Co-ordinates of the ends of the latus rectum `= (a, pm 2a)` `=(2, pm 4)` (ii) `y^(2) = -12x` Comparing with `y^(2)=-4ax` `4a = 12 rArr a=3` `:.` Co-ordiantes of vertex = (0,0) Co-ordiantes of focus `=(-a,0)=(-3,0)` Equation of axis `y=0` Equation of directrix `x = a rArr x =3` Length of latus rectum `= 4a =12` Co-ordinates of the ends of the latus rectum `= (-a, pm 2a)` `=(-3, pm 6)`. (iii) `x^(2)=6y` Comparing `x^(2)=4ay` `4a=6" "rArr" "a=(3)/(2)` `:.` Co-ordinates of vertex = (0,0). Co-ordainates of focus = `(0,a)=(0,(3)/(2))`. Equation of axis x=0. Equation of latus rectum y = - a `rArr" "y=-(3)/(2)`. Length of latus rectum = 4a=6. Co-ordinates of the ends of the latus rectum `(pm2a,a)=(pm3,(3)/(2))`. (iv) `x^(2)=-2y` Comparing with `x^(2)=-4ay` `4a=2" "rArr" "a=(1)/(2)` `:.` Co-ordinates of vertex = (0,0) Co-ordinates of focus `=(0,-a)=(0,-(1)/(2))` Equation of axis x = 0. Equation of directrix = a. `rArr" "y=(1)/(2)` Length of latus rectum = 4a = 2. Co-ordinates of the ends of latus ratus rectum `=(pm2a,-a)=(pm1,(1)/(2))`. |
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| 5. |
Find the co-ordinates of the points lying on parabola `x^(2)=12y` whose focal distance is 15. |
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Answer» Correct Answer - (12, 12), (-12,-4) |
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| 6. |
Find the equation of the hyperbola whose (i) vertices are `(pm3,0)` and foci are `(pm5,0)` (ii) vertices are `(p0,m2)` and foci are `(0,pm3)` (iii) Foci are `(0,pm5)` and length of transverse axis is 8. (iv) Foci are `(pm13,0)` and length of conjugate axis is 10. |
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Answer» Correct Answer - (i) `16x^(2)-9y^(2)=144` , (ii) `5y^(2)-4x^(2)=20` , (iii) `9y^(2)-16x^(2)=144` , (iv) `(x^(2))/(144)-(y^(2))/(25)=1` |
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| 7. |
Find the coordinates of the foci, the vertices, the eccentricity and the length of the latus rectum of the Hyperbola `(x^(2))/(16)-(y^(2))/(9)=1` |
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Answer» `(x^(2))/(16)-(y^(2))/(9)=1` Here, `a^(2)=16,b^(2)=9rArra=4,b=3` `:."Vertices"-=(pma,0)-=(pm4,0)` Eccentricity `e=sqrt(1+(b^(2))/(a^(2)))=sqrt(1+(9)/(16))=(5)/(4)` Coordinates of foci `-=(pmae,0)-=(pm5,0)` Latus rectum `=(2b^(2))/(a)=(2xx9)/(4)=(9)/(2)` |
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| 8. |
Find the coordinates of the foci, the vertices, the eccentricity and the length of the latus rectum of the Hyperbola `(y^(2))/(9)-(x^(2))/(27)=1` |
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Answer» `(y^(2))/(9)-(x^(2))/(27)=1` which is the equation of conjugate hyperbola Here, `b^(2)=9anda^(2)=27` `rArr" "b=3anda=3sqrt(3)` Coordinates of vertices `-=(0,pmb)-=(0,pm3)` Eccentricity `e=sqrt(1+(a^(2))/(b^(2)))=sqrt(1+(27)/(9))=2` Coordinates of foci `-=(0,pmbe)` `-=(0,pm3xx2)` `-=(0,pm6)` Length of latus rectum `=(2a^(2))/(b)=(2xx27)/(3)=18` |
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| 9. |
Find the vertices, transverse and conjugate axes, eccentricity, co-ordinates of foci, equation of directrices and length of latus rectum for each of the following hyperbola. (i) `(x^(2))/(16)+(y^(2))/(9)=1` (ii) `9x^(2)-4y^(2)=36` (iii) `16y^(2)-9x^(2)=576` (vi) `49x^(2)-16y^(2)=784` |
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Answer» Correct Answer - (i) Vertices `=(pm4,0), transverse axis = 8, conjugate axis = 6, `e=(5)/(4)`, foci `=(pm5,0)`, equation of directrices `x=pm(16)/(5)`, latus rectum `=(9)/(2)` (ii) Vertices `=(pm2,0)`, transverse axis = 12, conjugate axis = 16 , `(sqrt(13))/(2)`, foci `=(pmsqrt(13),0)`, equation of directrices `x=pm(4)/(sqrt(13))`, latus rectum =9 (iii) Vertices `=(0,pm6)`, transverse axis = 12 conjugate axis =16, `e=(5)/(3)`, foci = `(0,pm10)`, equation of directrices `y=pm(18)/(5)`, latus rectum `=(64)/(3)` (iv) Vertices `=(pm4,0)`, transverse axis = 8, conjugate axis = 14, `e=(sqrt(65))/(4),foci `=(pmsqrt(65),0)`, equation of directrices `x=pm(16)/(sqrt(65))`, latus rectum `=(49)/(2)`. |
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| 10. |
Find the vertices, eccentricity, foci, equation of directrices length of latus rectum for the hyperbola `3y^(2)-x^(2)=27`. |
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Answer» `3y^(2)-x^(2)=27`. `rArr" "(y^(2))/(9)-(x^(2))/(27)-1` which is the equation of conjugate hyperbola Here `b^(2)=9anda^(2)=27` `rArr""b=3anda=3sqrt(3)` Co-ordinates of vertices `(0,pmb)=(0,pm3)`. Eccentricity e `=sqrt(1+(a^(2))/(b^(2)))` `=sqrt(1+(27)/(9))=2`. Co-ordinates of foci `=(0,pmbe)` `=(0,pm3xx2)` `(0,pm6)`. Equation of directrices `y-pm(b)/(e)" "rArr" "y=pm(3)/(2)`. Length of latus rectum `=(2a^(2))/(b)=(2xx27)/(3)=18`. |
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| 11. |
Find the coordinates of the foci, the vertices, the eccentricity and the length of the latus rectum of the Hyperbola `49y^(2)-16x^(2)=784` |
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Answer» Equation of hyperbola `49y^(2)-16x^(2)=784` `rArr" "(y^(2))/(16)-(x^(2))/(49)=1` The transverse axis of hyperbola is along y-axis. Comparing with `(y^(2))/(b^(2))-(x^(2))/(a^(2))=1` `b^(2)=16," "a^(2)=49` `rArrb=4," "a=7` `:. "Vertices"-=(0,pmb)-=(0,pm4)` Eccentricity `e=sqrt(1+(a^(2))/(b^(2)))` `=sqrt(1+(49)/(16))=sqrt((65)/16)=sqrt((65)/4)` Now, `be4xx=(sqrt(65))/(4)=sqrt(65)` `:. "Coordinates of foci"-=(0,pmbe)-=(0,pmsqrt(65))` Length of latus latus rectum `=(2a^(2))/(b)=(2xx49)/(4)=(49)/(2)` |
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| 12. |
Find the coordinates of the foci, the vertices, the eccentricity and the length of the latus rectum of the Hyperbola `9y^(2)-4x^(2)=36` |
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Answer» Equation of hyperbola `9y^(2)-4x^(2)=36` `rArr(y^(2))/(4)-(x^(2))/(9)=1` Here, the transverse axis is along y-axis. Comparing with `(y^(2))/(b^(2))-(x^(2))/(a^(2))=1` `b^(2)=4," "a^(2)=9` `rArr" "b=2," "a=3` Now, vertices `-=(0,pmb)-=(0,pm2)` Eccentricity `e=sqrt(1+(a^(2))/(b^(2)))=sqrt(1+(9)/(4))` `=sqrt((13)/(4))=(sqrt(13))/(2)` Now, be `=2xx(sqrt(13))/(2)sqrt(13)` `:.` Coordinates of foci `-=(0,pmbe)-=(0,pmsqrt(13))` Length of latus rectum `=(2a^(2))/(b)=(2xx9)/(2)=9` |
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| 13. |
Find the coordinates of the foci, the vertices, the eccentricity and the length of the latus rectum of the Hyperbola `16x^(2)-9y^(2)=576` |
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Answer» Equation of hyperbola `16x^(2)-9y^(2)=576` `rArr(x^(2))/(36)-(y^(2))/(64)=1` `:.` The transverse axis of hyperbola is along x-axis. Comparing with `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` `a^(2)=36," "b^(2)=64` `rArr" "a=6," "b=8` `:. "Vertices" -=(pma,0)-=(pm6,0)` Eccentricity `e=sqrt(1+(b^(2))/(a^(2)))=sqrt(1+(64)/(36))` `=sqrt((100)/(36))=(10)/(6)=(5)/(3)` Now, ae `=6xx(5)/(3)=10` `:. "Coordinates of foci" -=(pmae,0)=(pm10,0)` Length of latus rectum `=(2b^(2))/(a)` `=(2xx(8)^(2))/(6)=(64)/(3)` |
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| 14. |
Find the vertices, eccentricity foci, rquation of directrices and length of latus rectum of the hyperbola `9x^(2)-25y^(2)=225`. |
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Answer» Equation of hyperbola. `9x^(2)-25y^(2)=225` `rArr" "(x^(2))/(25)-(y^(2))/(9)=1` Here `a^(2)=25andb^(2)=9` `rArr" "a=5andb=3` Co-ordinates of vertices `=(pma,0)=(pm5,0)`. Eccentricity e `=sqrt(1+(b^(2))/(a^(2)))=sqrt(1+(9)/(25))=sqrt(34)/(5)`. Co-ordinates of foci `=(pmae,0)=(pm5xx(sqrt(34))/(5),0)=(pmsqrt(34),0)`. Equation of directrices `x=pm(a)/(e)` `rArr""x=pm(5)/(sqrt(34)/(5))` `rArr""x=pm(5)/(sqrt(34))`. Length of latus rectum `=(2b^(2))/(a)=(2xx9)/(5)=(18)/(5)`. |
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| 15. |
Find the equation of a circle, the equation of whose two diameters are `x + y = 6 and 3x +4y =16` and radius is 10. |
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Answer» We know that the two diameters of a circle inersect at the centre of the circle `:.` The point of intersection `(8,-2)` of `x +y = 6 and 3x +4y =16` will be the centre of the circle. `:.` Equation of circle `(x-8)^(2)+(y+2)^(2) =10^(2)` `rArr x^(2)-16x+64+y^(2)+4y+4=100` `rArrx^(2)+y^(2)-16x+4y-32=0`. |
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| 16. |
Find the point on the parabola `y^(2)=18x` at which ordinate is 3 times its abscissa. |
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Answer» Correct Answer - (0,0) , (2,6) |
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| 17. |
The focal distance of a point of a parabola `y^(2)=8x` is 5. Find the abscissa of that point. |
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Answer» `y^(2)=8x` Comparing with `y^(2)=ax` 4a=8 `rArr" "a=2` Equation of directrix x=-a `rArr" "x=-2` `rArr" "x+2=0` Let `(x_(1),y_(1))` be any point on the parabola and its focal distance is 5. `:.` Distance of point `(x_(1),y_(1))` from directrix = distance of point `(x_(1),y_(1))` from focus `rArr" "x_(1)+2=5` `rArr" "x_(1)=3` `:.` Abscissa of required point = 3. |
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| 18. |
Find the equation of a line joining the vertex of parabola `y^(2)=8x` to its upper end of latus rectum. |
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Answer» `y^(2)=8x` Comparing with `y^(2)=4ax` 4a=8 a=2 Co-ordinates of the latus rectum L(a,2a)=L(2,4) Co-ordiantes of vertex A=(0,0) Equation of line joining the points A (0,0) and L(2,4) `y-0=(4-0)/(2-0)(x-0)` `rArr" "y=2x`. |
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| 19. |
The circles `x^(2)+y^(2)-2x-4y+1=0` and `x^(2)+y^(2)+4y-1 =0`A. `(1)/(a^(2))=(1)/(b^(2))+(1)/(c^(2))`B. `(1)/(b^(2))=(1)/(c^(2))+(1)/(a^(2))`C. `(1)/(c^(2))=(1)/(a^(2))+(1)/(b^(2))`D. None of these |
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Answer» Correct Answer - C |
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| 20. |
Find the equations of the circles the end points of whose diameter are as follows : (i) (0,6) and (6,0) (ii) (-1,3) and (2,4) (iii) (A+b,a-b) and (a-b,a+b) (iv) (0,0) and (2,2) |
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Answer» Correct Answer - (i) `x^(2)+y^(2)-6x-6y=0` , (ii) `x^(2)+y^(2)-4x-8y-214=0` (iii) `x^(2)+y^(2)-2ax-2ay+2(a^(2)-b^(2))=0` , (iv) `x^(2)+y^(2)-2x-2y=0` |
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| 21. |
If `e_(1)ande_(2)` be the eccentricities of the hyperbola `xy=c^(2)andx^(2)-y^(2)=c^(2)` respectively then :A. `e_(1)^(2)+e_(2)^(2)=1`B. `e_(1)^(2)+e_(2)^(2)=4`C. `e_(1)^(2)+e_(2)^(2)=6`D. `e_(1)^(2)+e_(2)^(2)=8`. |
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Answer» Correct Answer - B |
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| 22. |
Find the centre and radius of the circles`(x+5)^2+(y-3)^2=36` |
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Answer» `(x+5)^(2)+(y-3)^(2)=36` Comparing with `(x+5)^(2)+(y-k)^(2)=r^(2)` `h=-5,k=3,r^(2)=36` `rArr" "h=-5,k=3,r=6` `:.` Centre of circle `-=(h,k)-=(-5,3)` and radius = 6 units |
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| 23. |
Find the parametric equations of the circles `x^(2)+y^(2)=16`. |
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Answer» Given equation of circle `x^(2)+y^(2)=16` `rArr x^(2)+y^(2)=4^(2)` Let the parameter be `theta`, then Parametric equations `x =4 cos theta and y = 4 sin theta`. |
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| 24. |
Find the centre and radius of the circles`2x^2+2y^2-x=0` |
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Answer» `2x^(2)+2y^(2)-x=0` `rArrx^(2)+y^(2)-(1)/(2)x=0` Comparing with `x^(2)+y^(2)+2gx+2fy+c=0` `2g=-(1)/(2),2f=0,c=0` `rArr""g=-(1)/(4),f=0,c=0` `:. "Centre of circle"-=(-g,-f)-=((1)/(4),0)` and radius `=sqrt(g^(2)+f^(2)-c)` `=sqrt((-(1)/(4))^(2)+0-0=(1)/(4))` unit |
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| 25. |
Find the centre and radius of the circles`x^2+y^2-8x+10 y-12=0` |
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Answer» `x^(2)+y^(2)-8x+10y-12=0` Comparing with `x^(2)+y^(2)+gx+2fy+c=0` `2g=-8,2f=10,c=-12` `rArr" "g=-4,f=5,c=-12` `:.` Centre of circle `-=(-g,-f)-=(4,-5)` and radius `r=sqrt(g^(2)+f^(2)-c)` `=sqrt((4)^(2)+(-5)^(2)-(-12))` `sqrt(16+25+12)=sqrt(53)` units |
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| 26. |
Find the centre and radius of the circles`x^2+y^2-4x-8y-45=0` |
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Answer» `x^(2)+y^(2)-4x-8y-45=0` Comparing with `x^(2)+y^(2)+gx+2fy+c=0` `2g=-4,2f=-8,c=-45` `rArr" "g=-2,f=-4,c=-45` `:.` Centre of circle `-=(-g,-f)-=(2,4)` and radius `r=sqrt(g^(2)+f^(2)-c)` `=sqrt((2)^(2)+(4)^(2)-(-45))` `sqrt(4+16+45)=sqrt(65)` units |
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| 27. |
If `e_(1)ande_(2)` be the eccentricities of the ellipses `(x^(2))/(a^(2))+(4y^(2))/(b^(2))=1and(x^(2))/(a^(2))+(4y^(2))/(b^(2))=1` respectively then prove that `3=4e_(2)^(2)-e_(1)^(2)`. |
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Answer» For ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` Its eccentricity is `e_(1)` `:." "b^(2)=a^(2)(1-e_(1)^(2))rArr(b^(2))/(a^(2))=1-e_(1)^(2)` . . . (1) For ellipse `(x^(2))/(a^(2))+(4y^(2))/(b^(2))=1` `rArr" "(x^(2))/(a^(2))+(4y^(2))/(b^(2)//4)=1` Its eccentricity is `e_(2)` `:." "(b^(2))/(4)=a^(2)(1-e_(2)^(2))` `rArr" "(b^(2))/(4a^(2))=1-e_(2)^(2)" "rArr" "(b^(2))/(a^(2))=4-4e_(2)^(2)` . . . (2) From eqs. (1) and (2), we get `4-4e_(2)^(2)=1-e_(1)^(2)` `rArr" "3=4e_(2)^(2)-e_(1)^(2)`. |
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| 28. |
The ends of 20 cm long rod moves on two mutually perpendicular lines. If a point on this rod at 4 cm distance from end then from the eccentricity of the ellipse formed by moving that point. |
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Answer» Let the length of rod AB=20cm which moves on two mutually perpendicular lines OX and OY. Let point P(x,y) be at a didtance of 4 cm from end A. `:.` PB=AB-AP=20-4=16cm Let PM and PN are perpendiculars from p to x-axis and y-axis respectively. `:." "PM=yandPN-x` Let `anglePAM=theta" "rArr" "angleBPN=theta` in `DeltaAPM,sintheta=(y)/(4)and"in"DeltaBPN,costheta=(x)/(16)` Now `cos^(2)theta+sin^(2)=1` `rArr(x^(2))/(16^(2))+(y^(2))/(4^(2))=1`, which is the equation of the ellipse. Here a=16 and b=4 `:." "b^(2)=a^(2)(1-e^(2))` `"From "16=256(1-e^(2))` `rArr" "1-e^(2)=(1)/(16)` `rArr" "e^(2)=(15)/(16)` `rArr" "e=(sqrt(15))/(4)`. |
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| 29. |
Find the centre and radius radius of the following circles : (i) `(x-1)^(2)+(y+2)^(2)=16` (ii) `(x+2)^(2)+y^(2)=25` (iii) `(x+p)^(2)+(y+q)^(2)=p^(2)+q^(2)` (iv) `x^(2)+y^(2)=7` |
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Answer» Correct Answer - (i) C(1,-2), r=4 , (ii) C(-2,0), r=5 (iii) `C(-p,-q),r=sqrt(p^(2)+q^(2))` (iv) `C(0,0)r=sqrt(7)` |
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| 30. |
Prove that the curves x=4 `(costheta+sintheta)` and y=3 `(costheta-sintheta)` represents an ellipse. |
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Answer» `x=4(costheta+sintheta)` `rArr" "(x)/(4)=costheta+sintheta` . . . (1) and `y=3(costheta+sintheta)` `rArr" "(y)/(3)=costheta-sintheta` . . . (2) Squaring and adding eqs. (1) and (2) `(x^(2))/(16)+(y^(2))/(9)=(costheta+sintheta)^(2)+(costheta-sintheta)^(2)=2` `rArr" "9x^(2)+16y^(2)=288`. |
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| 31. |
Find the equation of circle whose : (i) radius is 5 and centre is (3,4). (ii) radius is `sqrt(5)` and centre is (0.2). (iii) radius is `sqrt(a^(2)+b^(2))` and centre is (a,b). (iv) radius is r and centre is `(rcostheta,rsintheta)`. (v) radius is `sqrt(a^(2)sec^(2)theta+b^(2)tan^(2)theta)` and centre is `(asectheta,btantheta)`. |
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Answer» Correct Answer - (i) `x^(2)+y^(2)-6-8y=0` , (ii) `x^(2)+y^(2)-4y-1=0` (iii) `x^(2)+y^(2)-2ar-2by=0` (iv) `x^(2)+y^(2)-2rcostheta*x-2rsintheta*y=0` (v) `x^(2)+y^(2)-2axsectheta-2bytantheta=0` |
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| 32. |
Find the equation of the ellipse whose centre is at origin, the distance between foci is 2 and eccentricity is `(1)/(sqrt(2))`. |
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Answer» Let the equation of the ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and agtb`. Here `e=(1)/(sqrt(2))` and the distance between foci = 2ae = 2 `rArr" "ae=1` `rArr" "a((1)/(sqrt(2)))=1` `rArr""a=sqrt(2)` `rArr" "a^(2)=2` and `b^(2)=a^(2)(1-e^(2))` `=a^(2)-(a^(2)e^(2))=2-1=1` `:.` Equation of ellipse `(x^(2))/(2)+(y^(2))/(1)=1` `rArr" "x^(2)+2y^(2)=2`. |
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| 33. |
Find the equation of hyperbola whose equation of directrix is x+2y=1, focus is (-1,-1) and eccentricity is `sqrt(2)`. |
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Answer» Correct Answer - `3x^(2)-3y^(2)-8xy+14x+18+8=0` |
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| 34. |
Find the equation of the circle with centre : (1, 1) and radius `sqrt(2)` |
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Answer» Centre `-=` (1,1) and radius = `sqrt(2)` units. `:.` Equation of circle, `(x-h)^(2)+(y-k)^(2)=r^(2)` `rArr" "(x-1)^(2)+(y-1)^(2)=(sqrt(2))^(2)` `rArr" " x^(2)-2x+1+y^(2)-2y+1=2` `rArr" "x^(2)+y^(2)-ax-2y=0` |
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| 35. |
Find the equation of the circle with centre :`(1/2,1/4)` and radius `1/(12)` |
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Answer» Centre `-=((1)/(2),(1)/(4))` and radius `(1)/(12)` unit. `:.` Equation of circle, `(x-h)^(2)+(y-k)^(2)=r^(2)` `rArr" "(x-(1)/(2))^(2)+(y-(1)/(4))^(2)=((1)/(12))^(2)` `rArr" "((2x-1)^(2))/(4)+((4y-1)^(2))/(16)=(1)/(144)` `rArr" "36(4x^(2)-4x+1)+(16y^(2)-8y+1)=1` `rArr" "144x^(2)-144x+36+144y^(2)-72y+9-1=0` `rArr" "144x^(2)+144y^(2)-144x-72y+44=0` `rArr" "36x^(2)+36y^(2)-36x-18y+11=0` |
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| 36. |
Find the equation of the circle with centre : `(-2, 3)`and radius 4 |
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Answer» Centre `-=` (-2,3) and radius = 4 units. `:.` Equation of circle, `(x-h)^(2)+(y-k)^(2)=r^(2)` `rArr" "(x-2)^(2)+(y-3)^(2)=4^(2)` `rArr" "x^(2)+4x+4=y^(2)-6y+9=16` `rArr" "x^(2)+y^(2)-4x-6y-3=0` |
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| 37. |
The ends of 20 cm rope are at two points 16 cm apart. Find the eccentricity and latus rectum of ellipse formed by the variablepoint at which the rope is tighten. |
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Answer» Correct Answer - `e=(4)/(5)`, latus rectum `=(36)/(5)` |
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| 38. |
Find the equation of the circle with centre :(0, 2) and radius 2 |
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Answer» Centre `-=` (0,2) and radius = 2 units `:.` Equation of circle, `(x-h)^(2)+(y-k)^(2)=r^(2)` `rArr" "(x-0)^(2)+(y-2)^(2)=2^(2)` `rArr" "x^(2)+y^(2)-4y+4=4` `rArr" "x^(2)+y^(2)-4y=0` |
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| 39. |
Convert the following equation of ellipse into standard from . (i) `16x^(2)+9y^(2)=144` (ii) `9x^(2)+25y^(2)=225` |
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Answer» Correct Answer - (i) `(x^(2))/(3^(2))+(y^(2))/(5^(2))=1`, (ii) `(x^(2))/(5^(2))+(5^(2))/(3^(2))=1` |
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| 40. |
(i) Find the equation of a circle passes through the origin and cuts the intercepts of 6 units and 8 units on X-axis and Y-axis respectively. (ii) Find the equation of a circle which passes through the two points on y-aixs whose distance from origin is 4 units and radius is 5 units. |
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Answer» Correct Answer - `x^(2)+y^(2)-6x-8y=0` |
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| 41. |
Find the equation of a circle which touches the Y-axis and whose centre is (3,0). |
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Answer» Correct Answer - `x^(2)+y^(2)-6x=0` |
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| 42. |
Find the equation of a circle which touches the X-axis and whose centre is `(asintheta,acostheta)`. |
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Answer» Correct Answer - `x^(2)+y^(2)-2axsintheta-2aycostheta+a^(2)sin^(2)theta=0` |
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| 43. |
Find the position of the point (1,-2) with respect to the circle `x^(2)+y^(2)+4x-2y-1=0`. |
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Answer» Correct Answer - External point |
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| 44. |
Find the equation of a circle passes through the origin and whose centre is (-2,5). |
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Answer» Correct Answer - `x^(2)+y^(2)+4x-10y=0` |
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| 45. |
A circle passes through `(0,0)` and `(1, 0)` and touches the circle `x^2 + y^2 = 9` then the centre of circle is -A. `((1)/(2),-sqrt(2))`B. (0,3)C. (-1,1)D. None of these |
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Answer» Correct Answer - A |
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| 46. |
Centre at (0,0) major axis on the y-axis passes through the points (3,2) and (1,6). |
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Answer» Centre is (0,0) Let equation of ellipse is : `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` . . . (1) `:.(3^(2))/(a^(2))+(2^(2))/(b^(2))=1rArr(9)/(a^(2))+(4)/(b^(2))=1` . . .(2) Ellipse passes through the point (1,6). `:.(1^(2))/(a^(2))+(6^(2))/(b^(2))=1rArr(1)/(a^(2))+(36)/(b^(2))=1` `rArr(9^(2))/(a^(2))+(324)/(b^(2))=9` . . . (3) Subtracting equation (2) from equation (3) `(320)/(b^(2))=8rArrb^(2)=40` Put the value of `b^(2)` in equation (2) `(9)/(a^(2))+(4)/(40)=1` `rArr(9)/(a^(2))=1-(1)/(10)=(9)/(10)rArra^(2)=10` From equation (1), equation jof ellipse `(x^(2))/(10)+(40)/(b^(2))=1` |
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| 47. |
Find the equation of the parabola whose is (2,-3) and directrix is x+4=0. |
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Answer» Let focus is S(2,-3) and P(x,y) be any point on the parabola `:.` PS=PM where PM=Length of perpendicular from P to directrix. `rArr" "sqrt((x-2)^(2)+(y-3)^(2))=x+4` `rArr" "(x-2)^(2)+(y-3)^(2)=(x+4)^(2)` `rArr" " x^(2)-4x+4+y^(2)+6y+9=x^(2)+8x+16` `rArr" "y^(2)+6y-12x=3` Which is the required equation of the parabola. |
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| 48. |
Find the equation of the circle passing throughthe points `(1,-2)a n d(4,-3)`and whose centre lies on the `3x+4y=7.` |
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Answer» Correct Answer - `x^(2)+y^(2)-6x+2y+5=0` |
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| 49. |
Find the equation of the circle with radius 5 whose centre lies onxaxis and passes through the point (2, 3). |
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Answer» Let the centre of circle on x-axis be (h,0). Given that radius of circle = 5 Therefore, equation of circle is `(x-h)^(2)+(y-0)^(2)=5^(2)` `rArr""(x-h)^(2)+y^(2)=25` . . .(1) This circle passes through the point (2,3). `:." "(2-h)^(2)+3^(3)=25` `rArr" "(2-h)^(2)=16` `rArr" "2-h=pm4` `rArr" "h=6orh=-2`. `:.` From equation (1) Equation of cicle `(x-6)^(2)+y^(2)=25or(x+2)^(2)+y^(2)=25` |
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| 50. |
Find the equation of parabola whose focus is (2,0) and equation of directrix is x=-2. |
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Answer» Equation of directrix x=-2 `rArr" "x+2=0` Let P(x, y) be any point on the parabola. Therefore, from the definition of parabola distance of from focus = perpendicular distance from p to directrix. `rArr" "sqrt((x-2)^(2)+(y-0)^(2))=x+2` `rArr" "(x-2)^(2)+y^(2)=(x+2)^(2)` `rArr" "y^(2)=8x`. |
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