Sr. Smart.pmd

FILE NO. 2435/5-03/10/15 SR.SMART & IC SMART MATHS T.TASK SOLUTIONS (05/10/15 TO 10/10/15)

I-ICON SPARK- JEE (MAIN) & EAMCET INTEGRATED STUDY MATERIAL 1

SR. SMART & IC SMART

ELLIPSEEAMCET PRACTICE QUESTIONS

7.a a4 4 2a 4e 1/ 2

a= 2 b2 = a2(1-e2)= 14 14

=3

Equation of the ellipse2 2

2 2x y 1 3x 4y 124 3

8. S12= 0 k = 69. Equation can be written as

2 2x 1 y 1 14 9

Equation of latus recta of

2 22 2

x y1

b a

are

y b ae y 1 5 10. The sides of a rectangle of greatest area that can be

inscribed in a ellipse 2 2

2 2

x y 1a b

are given equation

2 2x y 164 16

1 a 2 breadth b 2

8 2 4 2

11.144 36 3e

144 2

12. 9x2+5y2-18x-20y-16=0

2 2x 1 y 2 15 9

;2 2

2

b a 2eb 3

HYPERBOLA

26. eliminate ' '

x y x y 1.a b a b

2 2

2 2

x y 1a b

Hyperbola27. 2b = 5, 2ae = 13

b = 5/2 ae = 13/22 2 2 2b a e a

225 169 a4 4

12a2

28.2a2ae 3 2. e 3

e

29.2

0 b / atan 45ae

22

2

be e 1a

2e e 1 0

1 5e2



30. CA = AS

a = ae a

e =2, a = 7 2 2 2b a e 1 147 2 2x y 1

49 147

31.2x y x 1 x 1t; x 1

2 2 2

32.2 2

21 1100 x y 9(3x 4y 7)2 5

2 2 21 1 9 3x 4y 1x y 25

2 5 100 5

2 2 21 1 9 3x 4y 1x y2 5 4 5

1 1S ,2 5

, directrix is 3x+4y-7=0 latus rectum is

parallel to directrix and passing through the centre30x + 40y 23 =0

33. Let the tangent be x+y=a2 2x y 1

4 3

Tangency condition 2 2 2 2 2a b m n l4 3 = 2aa = 1

2c 1Area2 ab 2

Note: Change the 4th option as 12 .

34.1 1

1 2 2 21

2x ym mx a

2 2 1 44 3

35.

2 2x y 11 14 9

Tangent 2x sec 3y tan 1

2sec 3Tan 14 3 3

2sec3

1Tan63

36. Tangent to 2y 24x is

6y mxm

it is tangent of 2 25x y 5 2 2x y 1

1 5

2 2 2 2 2a b m n l 2

2

36m 5m

4 2m 5m 36 0 m 3 37. let P(x1, y1) be the pole

Equation of the polar is 1 12 2xx yy 1 0a b

it is tangent of 2 2

2 2

x y 1



2 22 21 1

4 4

x y 1a b

2 2 2 2

4 4

x y 1a b

38. Equation of the chord having (x1,y1) as its middle

points is 2 2

1 1` 1 1xx yy x y9 4 9 4

passes through

(1, 2)

then 2 2

1 1 1 1x 2y x y9 4 9 4

then locus of (x1,y1) is a hyperbola2 2x y x 2y

9 4 9 4

centre

2x 1 1 2y 2x & y 19 9 2 4 4

1 11cetre(x , y ) ,12

39. Given hyperbola is x (y +3) 4(y +3) = 25(x 4) y +3) = 25conjugate Hyp (x-4) (y+3) = 25

40. Given Hyperbola 2xy + 7x 6y 18 =0centre = (3, 3/2)Asymptotes 2xy + 7x 6y + k =0

Sub centre 33,

2

k 21

2xy 7x 6y 21 0 41. it is the rectangules Hyperbola

Angle b/w the asymptoles = / 2

42. centre f 0x

4x 3y 6 0

f 0y

3x 4y 13 0

Solving we get 3 14,

5 5

Since asymptotes of rectangular Hyp areperpendicualr Equation of the other asymptote

2x y k 0

3 142 k 05 5

4 k 0 k 4

43. 2 2 221 2 3x 4y 1x y 25

2 5 100 5

2 2 221 2 3x 4y 1x y2 5 4 5

2

14

2 or 2 44. length of the latusrectum of the hyperbola xy=c2 is

2 2c45. S(ae,0), S1 (-ae,0), A(a,0)

SA.S1A= a2e2-a2 = b2 = 12



64. 2 2

2 2

x h y ki 1

a b

centre = (h, k)(iv) Differentiate partially w.r.t x & y2x 4 =0 2y 2 =0x = 2 y +1 =0

y = 1

65. 2 2

2 2

x h y k1

a b

vertices h a,k

for 2 2

2 2

x h y k1

a b

vertices = h,k a

66. (i) 2 2

2

a b 144 36ea 144

180 5144 2

67. (i) 2 24x 9y 8x 4 0

2 24 x 2x 9y 40

2 24 x 1 9y 36

2 2x 1 y 19 4

Length of the transverse axis = 2b = 4

(ii) 2 29x 16y 18x 32y 151 0

2 2x 1 y 1 116 9

Lenght transverse axis = 2a = 8

68. for 2 2 2

2 2

x y 2b1, L.L.Ra b a

2 2 2

2 2

x y 2bfor 1, L.L.Ra b b

69. Apply S1 =0

70. Equation of the nomral 2 2

2 2

1 1

ax by a bx y

71. vertices of 2 2

2 2

c y1

a b

are

a,

72. 2 29x 16y 18x 64y 89 0

2 29 x 2x 1 16 y 4y 4 89 9 64

2 29 x 1 16 y 2 144

2 2x 1 y 2 116 9

L.conjugate axis = 2a = 2(4) =8

73.1 1

1 2 2 21

2x mm mx a

2 21

1 2 2 21

y bm mx a

74. Conceptual75. S12 = 076. Conceptual77. Conceptual

78. (i) x 2 cos t sin t



y 5 cos t sin t

2 2x y 24 25 (ellipse)

(ii) xy = 12 (rectangular hyperbola)

(iii) 2

2 yx 1 cot t 14

2y4x 4 2y 4 4x (parabola)

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