Practice Paper – 6 67 Senior Inter ✦ Mathematics – IIB SOLUTIONS PRACTICE PAPER – 6 SECTION – A I. I. I. I. I. 1. Find the value of k if the points (4, 2) and (k, –3) are conjugate points with respect to the circle x 2 + y 2 – 5x + 8y + 6 = 0. Sol. Equation of the circle is x 2 + y 2 – 5x + 8y + 6 = 0 Polar of P(4, 2) is x.4 + y.2 – 2 5 (x + 4) + 4 (y + 2) + 6 = 0 8x + 4y – 5x – 20 + 8y + 16 + 12 = 0 3x + 12y + 8 = 0 P(4, 2), Q(k, –3) are conjugate points Polar of P passes through Q ∴ 3k – 36 + 8 = 0 3k = 28 ⇒ k = 3 28 . 2. If the circle x 2 + y 2 + ax + by – 12 = 0 has the centre at (2, 3) then find a, b and the radius of the circle. Sol. Equation of the circle is x 2 + y 2 + ax + by – 12 = 0 Centre = − − 2 b , 2 a = (2, 3) 2 a − = 2 2 b − = 3 a = – 4 b = –6 g = –2, f = –3, c = –12 radius = c f g 2 2 − + = 12 9 4 + + = 5
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Practice Paper – 667
Senior Inter ✦ Mathematics – IIB
SOLUTIONS
PRACTICE PAPER – 6
SECTION – A
I.I.I.I.I. 1. Find the value of k if the points (4, 2) and (k, –3) are conjugatepoints with respect to the circle x2 + y2 – 5x + 8y + 6 = 0.
Sol. Equation of the circle is x2 + y2 – 5x + 8y + 6 = 0
Polar of P(4, 2) is
x.4 + y.2 – 25
(x + 4) + 4 (y + 2) + 6 = 0
8x + 4y – 5x – 20 + 8y + 16 + 12 = 0
3x + 12y + 8 = 0
P(4, 2), Q(k, –3) are conjugate points
Polar of P passes through Q
∴ 3k – 36 + 8 = 0
3k = 28 ⇒ k = 3
28.
2. If the circle x2 + y2 + ax + by – 12 = 0 has the centre at (2, 3)
then find a, b and the radius of the circle.
Sol. Equation of the circle is x2 + y2 + ax + by – 12 = 0
13. Find the equation of the tangent and normal to the ellipse9x2 + 16y2 = 144 at the end of the latus rectum in the firstquadrant. Mar. '07
Sol. Given ellipse is 9x2 + 16y2 = 144
9y
16x 22
+ =1
e = 2
22
a
ba − =
16916 −
= 47
End of the latus rectum in first quadrant
P
a
b,ae
2
=
49
,7
Equation of the tangent at P is
21
21
b
yy
a
xx+ = 1
+
49
.9y
167
.x = 1
4y
16x7 + = 1 or 7 x + 4y = 16
Equation of the normal at P is
1
2
1
2
yyb
xxa − = a2 – b2
−
49y9
7
x16 = 16 – 9
7
x16 – 4y = 7
16x – 4 7y = 7 7 .
Practice Paper – 674
Senior Inter ✦ Mathematics – IIB
14. If e, e1 are the eccentricities of a hyperbola and its conjugate
hyperbola prove that 21
2 e
1
e
1+ = 1.
Sol. Equation of the hyperbola is 2
2
2
2
b
y
a
x − = 1
∴ b2 = a2(e2 – 1) ⇒ e2 – 1= 2
2
a
b
e2 = 1 + 2
22
2
2
a
ba
a
b +=
∴ 22
2
2 ba
a
e
1
+= –––– (1)
Equation of the conjugate hyperbola is
2
2
2
2
b
y
a
x − = – 1 ⇒ 2
2
2
2
a
x
b
y− = 1
⇒ a2 = b2 (e12 – 1) ⇒ 2
1e – 1 = 2
2
b
a
21
e = 1 + 2
2
b
a = 2
22
b
ba +
22
2
21 ba
b
e
1
+= –––– (2)
Adding (1) and (2)
22
2
22
2
21
2 ba
b
ba
a
e
1
e
1
++
+=+ = 22
22
ba
ba
++
= 1.
15. Solvedx
4 + 5 sin x∫ Mar. '05
Sol. t = tan 2x
⇒ dt = sec2 2x
. 21
dx
dx = 22 t1
dt2
2x
sec
dt2
+=
Practice Paper – 675
Senior Inter ✦ Mathematics – IIB
sin x = 22 t1
t2
2x
tan1
2x
tan2
+=
+
I = ∫ ∫ ++=
++
+ t10t44
dt2
t1t2
54
t1
dt2
2
2
2
= ∫ ∫
−
+
=++
222
43
45
t
dt21
12t5
t
dt21
= C
43
45
t
43
45
tlog
43
.2
1.
21 +
++
−+
= C4t21t2
log31
C8t42t4
log31 +
++=+
++
= C2
2x
tan2
12x
tan2log
31 +
+
+
16. Evaluate /4
0
log (1+ tan x) dxπ
∫
Sol. I =π
π + − ∫�
��� ��� � ��
=π π −
+ π +
∫�
��� ��� ���� ��
��� ��� �
=π
−+ + ∫�
��� ���� ��
��� �
Practice Paper – 676
Senior Inter ✦ Mathematics – IIB
=
π + + − +
∫�
��� � ��� ���� ��
��� �
=π
− +∫�
���� � ��� � ��� �� ��
=π π
− +∫ ∫� �
��� � �� ��� � ��� �� � ��
= I−π 4/
0)x(2log
2I = 2log4π
I = 2log8π
17. Evaluate dxdy
= tan2 (x + y).
Sol.dxdy
= tan2 (x + y)
put v = x + y
dxdv
= 1 + dxdy
= 1 + tan2 v = sec2 v
∫ vsec
dv2 = ∫ dx
= ∫ cos2 v . dv = x + c
∫ +2
)v2cos1( dv = x + c
∫ (1 + cos 2v) dv = 2x + 2c
v + 2
v2sin = 2x + 2c
2v + sin 2v = 4x + c'
2(x + y) + sin 2(x + y) = 4x + c'
x – y – 21
sin [2(x + y)] = c
Practice Paper – 677
Senior Inter ✦ Mathematics – IIB
SECTION – C
III.III.III.III.III. 18. If (2, 0), (0, 1), (4, 5) and (0, c) are concyclic and then find c.
Sol. x2 + y2 + 2gx + 2fy + c1 = 0
Satisfies (2, 0), (0, 1) (4, 5) we get
4 + 0 + 4g + c1 = 0 –– (i)
0 + 1 + 2g. 0 + 2f + c1 = 0 –– (ii)
16 + 25 + 8g + 10f + c1 = 0 –– (iii)
(ii) – (i) we get
– 3 – 4g + 2f = 0
4g – 2f = – 3 –– (iv)
(ii) – (iii) we get
– 40 – 8g – 8f = 0 (or)
g + f = – 5 –– (v)
Solving(iv) and (v) we get
g = – 613
, f = – 617
Substituting g and f value in equation (i) we get
4 + 4
−
613 + c1 = 0
c1 = 3
14
Now equation x2 + y2 – 3
14y
317
x3
13 +− = 0
Now circle passes through (0, c) then
c2 –3
14c
317 + = 0
3c2 – 17c + 14 = 0
⇒ (3c – 14) (c – 1) = 0 (or)
c = 1 or 3
14 .
Practice Paper – 678
Senior Inter ✦ Mathematics – IIB
19. Show that the locus of the point of intersection of the linesx cos ααααα + y sin ααααα = a, x sin ααααα – y cos ααααα = b (ααααα is a parameter)is a circle.