Upload
aijazmona
View
231
Download
1
Tags:
Embed Size (px)
DESCRIPTION
important formula
Citation preview
1. Fundamental Relations between the Trigonometrical ratios of an angle
sin2 + cos2 = 1 or sin2 =1 –cos2 or cos2 = 1 – sin2
1 + tan2 = sec2 or sec2 – tan2 = 1
1 + cot2 = cosec2 or cosec2 – cot2 = 1
tan =
sin
coscotand
cos
sin
sin . cosec = tan . cot = cos . sec = 1
2 2
1 cotsin , cos ,
1 cot 1 cot
cot
cot1sec,cot1eccos
22
2. Trigonometric Ratios of Compound Angles
An angle made up of the algebraic sum of two or more angles is called compound angle.
Some formulae and results regarding compound angles:
sin (A + B) = sin A cosB + cosA sinB
sin(A – B) = sinA cosB – cos A sinB
cos (A + B) = cosA cos B – sinA sin B
cos(A – B) = cosA cosB + sin A sin B.
tan(A + B) = BtanAtan1
BtanAtan
, tan (45° + A) = Atan1
Atan1
tan(A–B) = BtanAtan1
BtanAtan
, tan (45° – A) = Atan1
Atan1
cot (A + B) = BcotAcot
1BcotAcot
, cot (A – B) = AcotBcot
1BcotAcot
sin(A + B) sin(A – B) = sin2A – sin2B = cos2B – cos2A
cos(A + B) cos(A – B) = cos2A – sin2B = cos2B –sin2A.
tan (A + B + C) = AtanCtanCtanBtanBtanAtan1
CtanBtanAtanCtanBtanAtan
3. Trigonometric Ratios of Multiples of an angle
sin2A = 2sinA cosA = Atan1
Atan22
cos2A = cos2 A –sin2A = 1 – 2 sin2A = 2 cos2A–1 = Atan1
Atan12
2
,
1 + cos2A = 2cos2A, 1 – cos2A = 2sin2A
1E
K GYAN
M
AN D I R
mv
TRIGONOMETRIC RATIOS & IDENTITIES
Maths Summer Vacation AssignmentPackage Solution
2E
K GYAN
M
AN D I R
mv
tan2AAtan1
Atan22
sin3A = 3sinA – 4sin3A = 4sin(60° – A) sinA sin(60° + A)
cos3A = 4 cos3A – 3cosA = 4cos(60°–A) cosA(cos60°+A)
Atan31
AtanAtan3A3tan
2
3
= tan(60°–A) tanA tan(60°+A)
4. Product of sines/cosines in term of sums
2 sinA cosB = sin (A + B) + sin (A – B) 2 cos A sin B = sin (A + B) – sin (A – B)
2 cos A cos B = cos (A + B) + cos (A – B) 2 sin A sin B = cos (A – B) – cos (A + B)
5. Sum of sines/cosines in term of products
sinC + sinD = 2sin 2
DCcos
2
DC sinC – sinD = 2 cos
2
DCsin
2
DC
cosC + cosD = 2cos 2
DCcos
2
DC cosC – cosD = –2sin
2
DCsin
2
DC
tanA + tanB = BcosAcos
)BA(sin , tanA – tanB =
BcosAcos
)BA(sin
cot tan 2cot 2A A A cot tan 2cosec2A A A
6. Maximum and minimum values of acos + bsin
– 2222 basinbcosaba
Hence the maximum value = 22 ba and minimum value is 22 ba .
7. Trigonometric Ratio of Submultiple of an Angle
Asin1|2
Acos
2
Asin|
or
otherwise,ve
4
3n2
2
A
4n2if,ve
Asin12
Acos
2
Asin
Asin1|2
Acos
2
Asin|
or
A 5ve, if 2n 2nA A
sin cos 1 sin A 4 2 42 2
ve, otherwise
Atan
11Atan
2
Atan
2 |a cosA + bsinA| 22 ba
Also cosA sinA = 2 sin
A
4 =
4Acos2
3E
K GYAN
M
AN D I R
mv
8. Conditional Identities : If A + B + C = , then
sin (B + C) = sinA, cosB = –cos (C + A)
cos (A + B) = –cosC, sinC = sin(A + B)
tan (C + A) = –tanB, cotA = –cot(B + C)
cos ,2
Csin
2
BA
2
BAsin
2
Ccos
2
CBcos
2
Asin,
2
Bcos
2
ACsin
2
ACcot
2
Btan,
2
Acot
2
CBtan
9. Some important identities:
If A, B, C are angles of a triangle (or A + B + C = ):
tanA + tanB + tanC = tanA tanB tanC
cotA cotB + cotB cotC + cotC cotA = 1
tan2
A tan
2
B + tan
2
B tan
2
C + tan 1
2
Atan
2
C
cot2
Ccot
2
Bcot
2
Acot
2
Ccot
2
Bcot
2
A
sin2A + sin2B + sin2C = 4sinA sinB sinC
cos2A + cos2B + cos2C = –1 – 4cosA cosB cosC
sinA + sinB + sinC = 4cos2
A cos
2
Ccos
2
B
cosA + cosB + cosC = 1 + 4 sin2
Csin
2
Bsin
2
A
10. Two Simple Trigonometrical Series
sin + sin( )+sin ( 2 ) + ... + sin{ )1n( }=
2sin
2
nsin
2
)1n(2sin
cos+cos( )+cos( 2 )+ ... +cos{ )1n( }=
2sin
2
nsin
2
)1n(2cos
4E
K GYAN
M
AN D I R
mv
1. (a)2 2cos cos cos cos
3 3x x x x
1 cos 2 1 cos(120 2 )cos cos(60 )
2 2
x xx x
2 cos(120 2 ) cos 2 2cos(60 )cos
2
x x x x
2 2cos(60 2 )cos 60 (cos(60 2 ) cos 60 )
2
x x
2 cos(60 2 ) cos(60 2 ) cos60
2
x x is independent of x
2. (b)1 2(2sin 70 sin10 ) 1 2(cos60 cos80 )
2sin10 2sin10
11 2 2cos(90 10 )
0 2sin1021
2sin10 2sin10
3.(d) (a) tan tan
180 180 tan(180 )1 tan tan
A BA B C A B C C
A B
tan tantan tan tan tan tan tan tan
1 tan tan
A BC A B C A B C
A B
tan tan tan tan tan tanA B C A B C
(b) 2 2 2 2 2( ) 5 3a b c ab a b c ab
2 2 2 3 3cos
2 2 2
a b cC
ab
which is not possible.
(c) sin : sin : sin 2 :3 : 7 : : 2 : 3 : 7 2 , 3 , 7A B C a b c a k b k c k
since c a b then : : 2 :3: 7a b c is not possible
(d) 3
cos cos sin sin2
A B A B cos (A – B) = 2
3A – B =
6
and
cos cos sin sin 0A B A B cos (A + B) = 0A + B = 2
Both imply that A = 60º, B = 30º.Hence such a triangle is possible
4. (c) 0 03 cos 20 sec 20ec
3 14 cos20 sin 20
2 23 1 2( 3 cos20 sin 20 )
sin 20 cos20 2sin 20 cos20 sin 40
4 cos30 cos 20 sin 30 sin 20 4cos(30 20 )4
sin(90 50 ) cos50
5E
K GYAN
M
AN D I R
mv
5. (a) cos68 cos52 cos(180 8 ) 120 16
2cos cos cos82 2
12 cos8 cos8 0
2
6. (c) )sin(sinsincossin2 2sin cos sin sin (sin cos cos sin )
Dividing by sin sin sin both side we get 2cot cot cot 2 1 1
tan tan tan
i.e. tan,tan and tan are in H.P....
7. (a) 2 9 4sin( ) 1 cos 1
25 5 2
2 16 3sin 1 cos 1
25 5 0
Now sin 2 sin sin cos cos sin
4 4 3 3 16 91
5 5 5 5 25 25
8.(c) 0
0
00
00
147cos
147sin
12sin12cos
12sin12cos
1 tan12tan(180 33 )
1 1.tan12
tan 45 tan12tan 33
1 tan 45 tan12
tan(45 12 ) tan33 tan33 tan33 0
9.(a) sin 600 cos330 cos120 sin150
sin(360 240 )cos(360 30 ) cos(180 60 )sin(180 30 )
sin 240 cos30 cos60 sin30
cos(360 ) cos ,cos(180 ) cos , sin(180 ) sin
sin(180 60 )cos30 cos60 sin 30
sin 60 cos30 cos60 sin30 sin(180 ) sin
sin(60 30 ) 1
10.(c) Given sin cosx y ..........(i) and 2 2sin (sin ) cos (cos ) sin cosx y
2 2sin (sin ) sin (cos ) sin cosx x by (i)
sin 1x sin cos cosx ....(ii)
Now by (i) and (ii) we get cos sin cos siny y ...(iii)
So by (ii) and (iii) we get 2 2 2 2cos sin 1x y
6E
K GYAN
M
AN D I R
mv
Useful cases of general solutions (where n Z )
1. (i) sin 0 n (ii) cos 0 (2 1) / 2n (iii) tan 0 n
2. (i) sin 1 2 ( / 2)n and sin 1 2 ( / 2)n
(ii) cos 1 2n and cos 1 (2 1)n
3. (i) sin sin cosec cosecor ( 1)nn
(ii) cos cos sec secor 2n
(iii) tan tan cot cotor n
4. For all: 2 2trig trig n
5. The equations of type acos sinb c are solved by transforming them to
cos cos where 2 2
cosa
a b
,
2 2sin
b
a b
and
2 2cos
c
a b
2 n
2n
which implies that solution exists if and only if 2 2 2 2a b c a b
Useful hints for solving trigonometric equations:
1. Factorize the equation using trigonometric formulae and identities. Each factor gives a part of solution.
2. Never cancel a common factor containing ' ' from the two sides of an equation.
For example, consider the equation tan 2 sin . If we divide both sides by sin , we get 1
cos2
,
which is clearly not equivalent to the given equation as the solutions obtained by sin 0 are lost. Thus,
instead of dividing an equation by a commong factor, take this factor out as a common factor from all termsof the equation.
3. Squaring should be avoided as far as possible. If squaring is done, check for extra solutions.
For example, consider the equation sin cos 1
On squaring, we get
1 sin 2 1 or sin 2 0 , 0, 1, 2,...2
nn
Clearly, and 3
2
do not satisfy the given equation, So, we get extra solutions. Thus, if squaring is
must, verify each of the solutions.
TRIGONOMETRIC EQUATIONS
7E
K GYAN
M
AN D I R
mv
2. (b) 4 2 2sin 2 2sin 0x x a 2 2 2 2(sin ) 2(sin ) ( 2) 0x x a
22 22 4 4( 2)
sin 1 32
ax a
.Now 20 1 3 1a
21 3 2a 21 3 4a 21 3 4a 22 1a
20 2 2 2a a
3. (a) 26
5cos
62sin
sin 2 1,
6
5cos 1
6
2 26 2
n
& 5
26
n
5
2 2 & 22 6 6
n n
6n
&
52
6n
7 13 19, , , ,........
6 6 6 6
&
7 19,
6 6
7 19, ,........
6 6
72
6n
4.(a) 2 24 10 ( 2) 6x x x i.e. minimum value of 2 4 10x x is 6
and max. value of 3sin 4cosx x is 2 23 4 5
i.e. graph of 3sin 4cosx x not intersect with 2 4 10x x i.e. no solution
5. (a)2 4 2 213 13
1 cos cos 1 cos (1 cos )16 16
2 2 131 sin cos
16
2 2 13 3sin cos 1
16 16
2
2 23 3(2sin cos ) sin 2 2
4 2 6n
9. (b) 21 cos sin 1x xy 2cos sin 2x xy 2cos 1x & sin 1xy
0x n & 3
22
xy n
3
, 22
x n yn
10. (b) 8sin cos cos 2 cos 4 sin 6 (sin 0)x x x x x x sin8 sin6 0 2cos7 sin 0x x x x
cos7 0 7 (2 1) (2 1)2 14
x x n x n
8E
K GYAN
M
AN D I R
mv
QUADRATIC EQUATION
1. The equation2 0ax bx c where 0, , ,a and a b c C (Complex numbers) is called
a quadratic equation. Its roots are
2 4
2
b b acx
a
2. If and are the roots of a quadratic equation, then
(a) Sum of the roots, /b a
(b) Product of roots, /c a
(c) Difference of the roots, | | | / |D a
(d) If roots are in the ratio m:n then
2 2( ) ( )m n
mn
(d) Equation formed by such given roots is
0x x or, 2 0x x
2x - (Sum of roots) x + (Product of roots) = 0
3. Nature of roots of 2 0ax bx c based on its Discriminant D =
2 4b ac :
If a, b, c R (Real numbers)
(i) D > 0 roots are real and distinct.
D 0 roots are real (may be equal or unequal)
D = 0 roots are real and identical.
D < 0 roots are non real complex (imaginary) conjugates eg. 2 + 3i, 2 - 3i.
Further If a, b, c Q (Rational numbers)
(i) D is a perfect square roots are rational.
(ii) D is not a perfect square roots are irrational conjugates eg. 2 + 3 , 2 - 3 .
4. Nature of roots of 2 0ax bx c based on the sign of a, b, and c:
when the sign of b matches with those of a and c, both roots are negative.
when the sign of b matches with that of c only, positive root is greater in magnitude.
Nature of roots Discriminant Sum of roots Product of roots
One +ve, one -ve D > 0 …… c/a<0
Both +ve roots D > 0 .-b/a > 0 c/a>0
Both -ve roots D > 0 .-b/a < 0 c/a>0
5. (a) Condition for one common root of two quadratic equations say
21 1 0a x b x c and 2
2 2 2 0a x b x c
Let the common root be , then 21 1 1 0a b c and 2
2 2 2 0a b c
Solving them by cross multiplication,
2
1 2 2 1 1 2 2 1 1 2 2 1
1
b c b c c a c a a b a b
9E
K GYAN
M
AN D I R
mv
The common root is 1 2 2 1 1 2 2 1
1 2 2 1 1 2 2 1
b c b c c a c a
c a c a a b a b
which is also the required condition.
(b) Condition for both common roots of two quadratic equations 2 0 1, 2i i ia x b x c i is
1 1 1
2 2 2
a b c
a b c
6. Sign of the quadratic expression 2ax bx c (a) If D < 0 then
(i) 2 0ax bx c for all x R when a > 0
(ii) 2 0ax bx c for all x R when a < 0
(b) If D > 0, sign scheme of y = ax2 + bx + c will be as follows ( , are roots of 2 0ax bx c )
Same as a Same as aOpposite in sign as that of a The above result can be understood by the different possible diagrams of
2y ax bx c , (shown below). The portion of the curve above the x-axis is positive and that below the x-
axis is negative
(a)
Y
O X
0, 0a D
(b)
Y
OX
0, 0a D
(c)
Y
OX
0, 0a D
(d)
Y
OX
0, 0a D (e)
Y
OX
0, 0a D (f)
Y
OX
0, 0a D
7. Maximum and Minimum value of a quadratic expression: From the above figure it is clear that,
(a) If a > 0, then y = 2ax bx c has a minimum value. It occurs at turning point (vertex) of the
parabola for which
24,
2 4 4
b ac b Dx y
a a a
. There is no maximum value.
(b) If a < 0, then 2y ax bx c has a maximum value. It also occurs at turning point of the curve
which is ,2 4
b D
a a
. There is no minimum value.
8. Useful conditions based on location of roots :
(a) Condition that both roots of f(x) = ax2 + bx + c = 0 will be greater than a number ‘d’ isD > 0, d < -b/2a and a.f(d) > 0
(b) Condition that both roots of f(x) = ax2 + bx + c = 0 will be less than a number ‘e’ isD > 0, e > -b/2a, a.f(e) > 0
(c) Condition that a number g lies between the roots of f(x) = 2 0ax bx c is
D > 0, a.f(g) < 0
10E
K GYAN
M
AN D I R
mv
(d) Condition that exactly one of the roots lies in the given interval (k1,k
2) is
D > 0, f(k1).f(k
2) < 0
(e) If sum of coefficients i.e. a + b + c = 0, then one root of 2 0ax bx c is 1.
9. Equation of more than two degrees :-
(a) If , , be the roots of 3 2 0ax bx cx d then
;b
a ;
b
a
d
a
(b) If , , , be the roots of 4 3 2 0ax bx cx dx c then
; ; ;b c d
a a a
e
a
(c) For a polynomial equation: 1 20 1 2 .... 0n n n
na x a x a x a 0a
Sum of roots =
n-1
n
Coeff.of x- ;
Coeff of x Product of roots =
n
n
Constant term-1 ;
Coeff. of x
10. Repeated roots :- If is n times repeated root of a polymial equation p(x) = 0, then n
p x x f x .
Hence, 0p , '' 0.....p , 1 0np .
11. Sign scheme of a polynomial function or rational function.
Let a < b < c < d and suppose
2x a x b
yx c x d
or, y = (x-a) (x-b)2 (x-c)(x-d)
(i) (x-b)2 never disturbs the sign of y as it is always positive (except at x = b).
(ii) y > 0 x (a, b) (b, c) (d, )
(iii) y < 0 x (- ,a) (c, d)
Now, we concise the above discussion in following steps :
Step I : Select the factor which disturbs the sign of y and find the value at which they become zero. (here, (x-a), (x-c), (x-d) disturb the sign and become zero at x = a, c, d)
Step II : Show these values on a number line
d
Step III : Check the sign of y for a specific interval and sign of subsequent intervals occurs alternately.
Here for x > d, y is +ve hence SIGN SCHEME of y will be as follows :
d vcve
(Note that at b (i.e. at a point between a and c, y = 0 which is neither +ve nor -ve and be careful about it whilewriting the final answer)
11E
K GYAN
M
AN D I R
mv
1. (a) Let 2( )f x ax bx c ,4a + 2b + c = 0 (2) 0f .Also 0b
a sum of roots 0
for 2 we get 2 i.e. both the roots are real.
2. (b) Let 2( ) 6f x ax bx then (0) 0f . Since ( ) 0f x has no distinct real roots then
(2) 0 4 2 6 2 3f a b a b then the least value of 2a + b is –3.
3. (b)2 4
2
b b acx
a
. Now 0
b
a , 2 24 4 0b ac b b b ac and
2 4b ac may be 0 or 0 then roots are with negative real parts.
4. (a) 1 ,a b c are of same sign then 0, 0 0, 0b c
a a & 2 4 0b ac then both the roots
are positive , if both C1 and C
2 are satisfied.
5. (c) 2 2 2 2 24 ( 1) 4 ( 2) 2 1 4 8 3 6 1b ac n n n n n n n n n
2 21 3( 2 ) 4 3( 1)n n n is perfect square for 2, 1,0n .
6. (b) Since a, b, c are distinct then 2 2 2 2 2 21( ) ( ) ( ) 0
2a b c ab bc ca a b b c c a then
a3 + b3 + c3 = 3abc 0a b c .Also since roots of equation 2 0cx ax b are equal then root of the
quadratic equation is 1.
7. (d) 2 25 5 1 5 4 0 ( 1)( 4) 0 1 4
2 2 5 52 3 4 1 2 3 5 0 ( 1) 0 1
2 2
.......(ii)
By (i) & (ii) 5
12
.Now 22 25 5 4 2 3 4D .Now for
2 22 3 4, 4 9 32 0b ac then 22 3 4 0 0D for 5
12
.So, 5
1,2
10.(c) 2 22 2 0x y y x x y x y .Now 2 24 4 0 1 0 1 1y y y
Now
2
2 1 92
2 4y y y
i.e.
9,0
4y
12E
K GYAN
M
AN D I R
mv
COMPLEX NUMBER
1. Definition of a complex number : A number of the form z x iy , where ,x y R and 2 1i is
called a complex number.
Re( )x z , is called real part of z and Im( )y z , is called imaginary part of z.
z x iy is called
(a) real if 0y (b) imaginary if 0y (c) purely imaginary if 0x
On this basis ‘0’ is a real and purely imaginary number as well.
2. Equality of two complex number :
(a) x iy a ib x a and y b i.e. real and imaginary parts are separately equal.
(b) Inequality does not hold in a complex plane. i.e. 3 4 2 3i i has no sense.
3. Representation of complex number :
(a) Algebraic form : z x iy .
(b) Ordered pair form : z = (x,y). z is also represented on a plane as apoint (x,y). Its real part is shown on the x-axis (real axis) and imaginarypart on the y-axis (imaginary axis). This plane is known as Complexplane or Argand plane or Gaussian plane.
(C) Polar form or trigonometric form : z = r (cos + isin ) where r = |z|
and = arg z.
(e) Euler or Exponential : iz re
4. Three basic terms of a complex number
(a) Modulus: If ,z x iy its modulus, is denoted by 2 2| | 0r z x y
Geometrically, it is the distance of point z from the origin.
(b) Argument (or amplitude) :
Geometrically, it is angle made by the line joining the origin to the point z(x, y), with x-axis.
If z = x + iy, then
1arg( ) tan /z y x ( z being in first quadrant)
1arg( ) tan /z y x ( z being in second quadrant)
1arg( ) tan /z y x ( z being in third quadrant)
1arg( ) tan /z y x ( z being in fourth quadrant)
(c) Conjugate of a complex number : If z = x + iy, then its conjugate is denoted by z x iy .
Geometrically, it is the reflection of point z with respect to x-axis.
O X
,x y
Real axis
PY
13E
K GYAN
M
AN D I R
mv
5. Properties of modulus, argument and conjugate :-
i. 1 2 1 2| | | || |z z z z ; | | | |n nz z ;11
2 2
zz
z z
ii. Triangular Inequalities:
1 2 1 2 1 2|| | | || | | | | | |z z z z z z and 1 2 1 2 1 2|| | | || | | | | | |z z z z z z
Note: Equalities 1 2 1 2| | | | | |z z z z and 1 2 1 2| | | | | |z z z z hold if and only if 1z and 2z are
collinear and are on the same side of the origin.
iii. | | | |,z z z z ;1 2 1 2z z z z ;
1 2 1 2z z z z ;1 1
2 2
z z
z z
; 2| |zz z
iv. Re( )2
z zz
; Im( )
2
z zz
i
; | | | |z Re z z ; | | Im | |z z z
If ,z z then the complex number is real. If ,z z then it is imaginary..
v. 2 2 21 2 1 2 1 2| | | | | | 2 Re( )z z z z z z or 2 2 2
1 2 1 2 1 2 1 2| | | | | | 2 | || | cos arg argz z z z z z z z
vi. 1 2 1 2arg arg argz z z z ; arg argnz n z ; 1 2 1 2arg / arg argz z z z
vii. arg argz z ; arg arg 0kz z if k and arg arg 0kz z if k
6. Cube roots of unity :
(a) Solving 3 1x , we get:1 3 1 3
1, ,2 2
i ix
; called cube root of unity, denoted by 1, &
2 .
(b) Properties of cube roots of unity
i. Their sum i.e. 21 0 and their product i.e. 3 1 ii. If we square one of the complex cube root of unity, we get the other.
iii. Their modulus i.e. 2| | | | 1 ; their arguments are 0, 2 / 3 and 2 / 3 .
iv. 21 3n n or 0 according as ‘n’ is a multiple of 3 or not.
v Cube roots of unity are vertices of an equilateral traingle which is inscribed in a circle of unit radius with centre at origin.
7. De Moivre’s Theorem : To find rational powers of a complex number (Polar form is must)
(a) If n is an integer then (cos sin ) cos sinni n i n
(b) If n is an integer then 1/ 2 2
(cos sin ) cos sinn k ki i
n n
where k=0,1,2,......,n-1.
8. Euler’s theorem : cos sin ii e ,
Hence, we have
(a) (cos sin )(cos sin ) /(cos sin )A i A B i B C i C cos sinA B C i A B C
(b) 1
cos sincos sin
ii
14E
K GYAN
M
AN D I R
mv
9. nth roots of unity :
i. If we consider
2i
ne
, then nth roots are 2 3 11, , , ,......., n .
So nth roots of unity are in G.P. ii. Sum of n roots of unity = 0 , and
Product of n roots of unity = 1( 1)n
Sum of nth roots of a complex number z0
= 0;Product of the nth roots of z
0 = ( -1)n+1 z
0
iii. Modulus of each of n roots = 1; their argument
0, 2 / , 4 / , ......,[2( 1) / ]n n n n are in A.P..
iv. These n roots of unity lie on a unit circle as vertices of a regular polygon of n sides.10. Square roots of a complex number
(i) 1/ 2 2 2
0 0 0 0, 2x iy x iy x y x xy y . Solve for x and y..
(ii) 1/ 2
cos sino o or i
1
2 cos sin2 2o o
or i
. Here, use tan( / 2) (1 cos ) / sin ;
11. Geometrical relations
(a) If 1z and 2z are two complex numbers, then the complex number 1 2nz mz
zm n
divides the join of 1z
and 2z in the ratio :m n .
(b) 1 2z-z z-z is equation of perpendicular bisector of line joining the points 1z and z2.
(c) Different forms of “Equation of a circle” are
i. 0| | ,z z r represents a circle with radius ‘r’ and centre 0z .
ii. 0zz az az b , (where ‘b’ is a real number), represents a circle with centre ‘-a’ and radius ‘2
a b
iii. 1
2
arg ( )z z
z z
represents arc of a circle through z
1 and z
2.
iv. 1 2z-z z-z ( 1)k if k is also a circle. Ends of diameter (k : 1) and center (- k2 :1).
(d) The triangle whose vertices are represented by 1 2 3, ,z z z is equilateral if and only if
2 3 3 1 1 2
1 1 10
z z z z z z
or 2 2 2
1 2 3 1 2 2 3 3 1z z z z z z z z z
12. Concept of rotation :
(a) Complex number ize is obtained by rotating Oz with an angle in
anticlockwise direction. It follows directly from the geometrical meaning ofmultiplication of two complex number.(b) In general, we have
3 13 1
2 1 2 1
cos sinz zz z
iz z z z
2(2 / )ne
(1,0)1n
( 1,0)
O1z
2z
3z
15E
K GYAN
M
AN D I R
mv
1.(a) 50
1 i can be written as 25 252 21 1 2i i i
25 25 25 25 252 2 2 0 2i i i i .
Hence real part of 50
1 0i
2. (b)2 1 1 4(5 5 ) 1 21 20
(5 5 ) 02(1) 2
i iz z i z
. Now 21 20 25 4 20i i
2 225 2 2 5 2 (5 2 )i i i then
1 (5 2 ) 6 2
2 2
i iz
or
4 23
2
ii
or 2 i .Then the product of the real part of the roots of z2 – z = 5 – 5i is (– 6).
3.(c) 2 0z z 2
0x iy x iy (if z x iy ) 2 2 2 0x y xyi x iy
2 2 2 0x y x xy y i .By equality of complex no. 2 2 0x y x .... (1)
and 2 0 2 1 0xy y y x 0y or 1
2x
.
Put 0y in (1) we get 2 0 ( 1) 0x x x x 0, 1x
and put 1
2x
in (1) we get
21 10
4 2y i.e. 2 1
4y which is not posible since y is real.
hence, 0 0z and 1 0i i.e. number of solutions is 2.
4.(c) Given equation 2 ( ) 3 0x p iq x i .If & are the roots then p iq & 3i .
Given 2 2 8 2
2 8 2( ) 6 8p iq i 2 2 (2 6) 8p q i pq
comparing real & imaginary parts, 2 2 8p q & 3pq .On solving we get 3& 1p q
or 3& 1p q
5.(a) It is given that for a complex no. z x iy
then Re Im 1z z represents 1x y
1x y- + =
1x y+ =
1x y- =1x y- - =
which is the equation of square.
6.(c)
3 4 25 3 425 25. 3 4
3 4 3 4 3 4 9 16
i ii
i i i
Now by figure we see that opposite angles are supplementary.
(1,0)
(3,4)(–3,4)
(–1,0)
2 52 5
Hence it is a cyclic quadrilateral.
7.(d) For a complex no. z it is given that 3z z (i).Takiing modulus of both sideswe get3
z z
21 0z z ( | | | |)z z 0z or
21z 0z or 1zz 0z or 4 1z by (i)
Hence 0; 1,z z z i .Hence no. of roots are 5.
16E
K GYAN
M
AN D I R
mv
8.(c) Let F is x iy then reciprocal of F is 2 2
1 x iy
x iy x y
i.e. reciprocal number is canjugate and its
modulus is less then 1 since modulus of F is greater than 1 i.e. C is required number..
9.(a) we have, 5
15
z i
z i
| 5 | | ( 5 ) |z i z i z is equidistant from 5i & -5i
z lies on the perpendicular bisector of line segment joining (0,5) & (0, 5) i.e. x -axis
10. (c) 2 2 21 2 1 2 1 2| | | | | | | || |z z z z z z 2 2 2 2
1 2 1 2 1 2 1 2 1 2| | | | 2 | || | cos( ) | | | | 2 | || |z z z z z z z z
1 2 1 2 1 2cos( ) 1 0 i.e. Arg z1 Arg z2 = 0
17E
K GYAN
M
AN D I R
mv
STRAIGHT LINE
1. Coordinate Systems: Cartesian and Polar Systems
(a) x = r cos , y = rsin
(b) 222 yxr , = tan-1(y/x)
(c) x coordinate is also called abscissa and y coordinate is calledordinate.
2. Distance between two points (x1, y
1) and (x
2, y
2) is
2 2
2 1 1 2x x y y
3. Section Formula: If a point P divides the line segment joining points (x1, y
1) and (x
2, y
2) in the ratio
(a) m : n internally, then2 1 2 1,
mx nx my nyP
m n m n
(b) m : n externally, then2 1 2 1,
mx nx my nyP
m n m n
(c) If P is the mid point, then 1 2 1 2,
2 2
x x y yP
(d) The ratio in which straight line 0 cbyax divides the line segment joining the points 11, yx and
22, yx is = 1 1
2 2
ax by c
ax by c
. Thus, points (x1, y
1) and (x
2, y
2) are on the same side (or opposite sides) of
the line 0ax by c according as 1 1ax by c and 2 2ax by c have same sign (or opposite sign).
4. (a) Area of a triangle whose vertices are (x1, y
1), (x
1, y
2) and (x
3, y
3) is
2 21 1 3 3
3 32 2 1 1
1
2
x yx y x y
x yx y x y =
1 1
2 2
3 3
11
12
1
x y
x y
x y
(b) Area of a polygon whose consecutive vertices are (xi, y
i) (i = 1, 2, 3, ...n) is
2 2 1 11 1
3 32 2 1 1
1......
2
n n n n
n n
x y x yx y x y
x y x yx y x y
0
,, roryxP
X
Y
1P
2P
( , )P x ym
n
18E
K GYAN
M
AN D I R
mv
5. (a) Centroid : The point (G) where medians of the triangle meet.We must note that
(i)2
1
AG BG CG
GD GE GF
(ii)1 2 3 1 2 3,
3 3
x x x y y yG
(b) Incentre : is the point (I) where internal bisectors of the angles meet.We must note that
(i)BD AB c
DC AC b etc.
(ii)1 2 3 1 2 3,
ax bx cx ay by cyI
a b c a b c
(c) Excentres : are points where angle bisectors of one internal andtwo external angles meet.
Excentre opposite to A 1 2 3 1 2 3,
ax bx cx ay by cy
a b c a b c
etc.
(d) Circumcentre : The Point (O) where perpendicular bisector of the sides ofthe triangle meet. It is centre of the circle passing through the vertices.(i) AO = BO = CO = R, called circumradius
(ii) ACODBOD (iii) Slope of OD x Slope of BC = -1 etc.
(e) Orthocentre : is the point (P) where altitudes of the triangle meet.We must note that :
Slope of AD x Slope of BC = -1 etc.
6. Properties of centres :(a) In an equilaterial triangle, the four centres (centroid, incentre, circumcentre and orthocentre) are
coincident.
(b) In a right angled triangle ABC 090A orthocentre is at A and circumcentre is the mid point of
hypotenuse BC.(c) Lines joining the centroid and vertices of a triangle ABC, divide the triangle into three equal areas.(d) Orthocentre (P) centroid (G) and circumcentre (O) of a triangle are collinear and
also, 1
AP PG AG
OD GO GD
O
GP
A
B CD
7. Slope of a line segment : If a line makes an angle with x-axis in anticlockwise direction, then tan is
called slope of the line. It is generally denoted by ‘m’.
Instead of x-axis, if the line makes an angle with positive direction of y-axis , then slope = tan2
Slope of a line joining the points (x1, y
1) and (x
2, y
2) is =
2 1
2 1
y y
x x
.
G
B D C
F E
33, yx 22 , yx
11, yxA
1 1,A x y
FE
P
DB C 2 2,x y 3 3,x y
19E
K GYAN
M
AN D I R
mv
8. Collinearity of points A, B, and C:(a) Slope of AB = Slope of BC (= Slope of CA)(b) Area of triangle ABC = 0(c) A divides BC in some ratio, i.e. Section formula holds.(d) Sum of two of AB, BC, and AC is equal to the third.
9. Locus : If a point moves such that it follows some (geometrical) condition, then the path traced out by thepoint is called its locus, and mathematical relation thus obtained is called equation of the locus.It is generally an equation connecting the coordinates (x and y) of the point and the given constants.
10. Trasformation of Coordinate System:General equations of transformation of coordinate system: On shifting of origin to point (h,k) and rotation of
axes by an angle , the old coordinates (x,y) and new coordinates (x’,y’) of a point are related by:
'cos 'sin
'sin 'cos
x x y h
y x y k
; which become '
'
x x h
y y k
in case of only translation of axes (i.e. no rotation)
11. Various useful forms of a straight line: Equation of a straight line which is
(a) Parallel to x-axis is y = c (a constant)Parallel to y-axis is x = c (a constant)Eq. of x-axis : y = 0; and that of y-axis : x = 0.
(b) Passing through origin, having slope m, is y = mx(c) Having slope m and intercept on y-axis c is y = mx+c(d) Having slope m and passing through (x
1, y
1) is y - mx = y
1 - mx
1
(e) Passing through points (x1, y
1) and (x
2, y
2) is 2 1
1 1
2 1
y yy y x x
x x
(f) Having intercepts a and b on the axes respectively is 1b
y
a
x
(g) If length of perpendicular from the origin to a line is p and the
perpendicular line (OP) makes an angle with the positive
direction of x-axis, then the equation of the straight line (AB) is cos sinx y p
(h) Distance form or paramatric form of a straight line is1 1
cos sin
x x y yr
A point at a distance “r” from the point (x1, y
1) on the
straight line can be taken as 1 1, cos , sinx y r x r y
(i) Equations of straight lines (PA and PB in the figure) passing
through the point (x1, y
1) and makin an angle with the
straight line of slope m tan are 1 1tany y x x
For 11(g) For 11(h) For 11(i)
A
B
C
20E
K GYAN
M
AN D I R
mv
12. General equation of a straight line is ax+by+c=0. Its slope = -a/b, and its distance from origin = 22 ba
c
13. Angle between two straight lines: whose slopes are m1 and m
2 is, 1 2 1
1 2
tan1
m m
m m
Acute and obtuse angle between them are 1 2 1
1 2
tan1
m m
m m
and
1 2 1
1 2
tan1
m m
m m
respectively..
If one line is parallel to y-axis and slope of the other line is ‘m’ then angle between them is 1tan
2m
14. Parallelism and Perpendicularity: Two straight line y = m1x + c
1 and y = m
2x + c
2 are
(a) parallel if m1 = m
2 slopes are equal
(b) perpendicular if m1m
2 = -1 i.e. product of slopes = -1.
15. If aix + b
iy + c
i = 0 (i = 1, 2) are two lines and
(a) 1 1
2 2
a b
a b lines are intersecting. (b)
1 1 1
2 2 2
a b c
a b c lines are parallel.
(c) 1 1 1
2 2 2
a b c
a b c lines are coincident. (d) 1 2 1 2 0a a b b lines are perpendicular..
16. (a) Equation of a straight line parallel to the line 0ax by c can be taken as 0ax by k .
(b) Equation of a straight line perpendicular to the line 0ax by c can be taken as 0bx ay k .
17. Perpendicular distance of point (x1, y
1) from the line 0ax by c is
1 1 1
2 2
ax by c
a b
Distance between two parallel lines 0ax by c and 0ax by d is 2 2
d c
a b
18. Concurrency: Three lines 0i i i iL a x b y c (i =1, 2, 3) are concurrent if
(a) Point of intersection of any two lines satisfies the third one, or
(b)
1 1 1
2 2 2
3 3 3
0
a b c
a b c
a b c
19. Equation of a family of lines (or a variable line) passing through the point of intersection of the lines
L1 = a
1x + b
1y + c
1 = 0 and L
2 = a
2x + b
2y
+
c
2 = 0 is 1 2 0L L
20. Equation of bisectors of the angle between the straight lines
0111 cybxa and 0222 cybxa are
1 1 1 2 2 2
2 2 2 21 1 2 2
a x b y c a x b y c
a b a b
If the signs of 1 2andc c are kept same then for origin containing angle bisector, " " sign is taken; and for acute
angle bisector, sign opposite to that of 1 2 1 2" ''a a b b is taken.
21. Foot of perpendicular from (x0,y
0) to the line ax+by+c = 0 is given by:
2 2
o oo oax by cx x y y
a b a b
.
21E
K GYAN
M
AN D I R
mv
1. Question incomplete
2. (c) Slope of 2 0 1
0 4 2AB
and
2 2
2 0 0 4 2 5AB
then 5BC . Since D is centre of square D
B(0,2) 45°55
(2,1)C
O (4,0)A
then slope of 2CD and 5CD
coordinate of C is 4 0 0 2
, (2,1)2 2
Now equation of CD is 2 1
51 2
5 5
x y coordinate of D is 3, 3x y
4. (a) Slope of 2
1AB
x
then
2tan
1ABX
x
A(1,2)
C (5,3)
B ( ,0)xX' Xslope of
3
5BC
x
then
3tan
5CBX
x
Now 2 3
10 2 3 3 5 131 5
x x xx x
.Now slope of
2 10 5
13 8 415
AB
So equation of AB is 5 4 13x y .
5.(c) Equation of the median through A is (px + qy -1) (qx + py - 1)=0.Since it passes through (p, q) then
2 2 1
2 1
p q
pq
then the equation of the median is(2pq 1)(px + qy 1) = (p2 + q2 1) (qx + py1).
6.(b) Coordinates of point A is (3,0) .Coordinate of point B is (0,4) .Coordinate of point P is 21 72
,25 25
Now slope of PA is 72 3
96 4
,slope of PB is 28 4
21 3
.Since
3 41
4 3
then PA PB then AB is diameter of PAB i.e. circumradius 5
2
22E
K GYAN
M
AN D I R
mv
7.(a) Let AB r then equation of line AB iscos sin
x p y qr
cos , sinx p r y q r will lie in
0ax by c then cos sin 0ap ar bq br c
cos sin
ap bq cr
a b
AB
ap bq c
a b
cos sin
8. (a) If origin is shifted to ( , )h k then new coordinates of ( , )x y is ( , )x h y k . So the triangle formed by
the points (x1 + h, y
1 + k) , (x
2 + h, y
2 + k) and (x
3 + h, y
3 + k) is congruent to the triangle formed by the
points (x1, y
1), (x
2, y
2) and (x
3, y
3) then 'A A .
10.(a) Equation of bisectors of angle formed by k1u k
2v = 0 and k
1u + k
2v = 0 for non zero real k
1 and k
2 are
1 2 1 2
2 2 2 21 2 1 2
k u k v k u k v
k k k k
or 1 2
2 21 2
k u k v
k k
1 0k u or 2 0k v 0, 0u v
then equation of the bisectors of angle formed by is 0uv
23E
K GYAN
M
AN D I R
mv
CIRCLE
1. Circle is locus of a moving point such that its distance from a fixed point remains fixed.
General equation of second degree in x and y, 0222 22 cfygxbyhxyax represents a circle if
a = b and h = 0.
2. Equation of circle in various forms :
(i) Central form: 2 2 2x h y k r has centre (h, k) and radius r..
eg. circle with centre ,h k and touching x-axis is 2 2 2x h y k k
circle with centre ,h k and touching y-axis is 2 2 2x h y k h
(ii) General equation: 2 2 2 2 0x y gx fy c has centre (-g, -f) and radius 2 2g f c
the circle will be a real circle, point circle or imaginary circle depending on whether
2 2 0, 0g f c or 0 ;
the circle passes through the origin if 0c ;
Length of the intercept made by the circle 2 2 2 2 0x y gx fy c on x-axis is 22 g c and that
on y-axis 22 f c .
(iii) Diametric form: Equation of the circle having 11, yx and 22, yx as extremities of a diameter is
1 2 1 2 0x x x x y y y y
(iv) Parametric form:
For circle 2 2 2x y r : cosx r , siny r
For 2 2 2x h y k r : cosx h r , siny k r
Parameter is such that .
A(x )1, y1
(x,y)B(x y )2, 2
24E
K GYAN
M
AN D I R
mv
3. Position of a point w.r.t. a Circle:
(i) Condition that point 11, yx lies inside, outside or on the circle 2 2 2 2 0S x y gx fy c
is 2 21 1 1 1 12 2 0, 0, 0S x y gx fy c respectively..
(ii) The greatest and least distance of a point P from a circle with centre C and radius r is PC r and
| |PC r
(iii) For a point on the circle: Equation of the tangent at any point (x1,y
1) on the circle S=0 is T = 0
(where 1 1 1 1T xx yy g x x f y y c ). This is called as Point form of tangent.
(iv) For a point inside the circle: Eq. of chord of a circle S = 0, whose middle point is (x1, y
1) is T = S
1.
(v) For a point outside the circle:
(a) Length of tangent from a point outside the circle to the circle S = 0 is 1S . Here, the square of the
length of tangent, i.e. S1, is called the Power of this point.
(b) Angle between these tangents 112 tan /r S
(c) If two tangents PT1 and PT
2 are drawn from point P to a circle S = 0 then the equation of pair of
tangents is SS1 = T2.
(d) The equation of chord of contact (T1T
2) of tangents from the outside point (x
1, y
1) to the circle S=0 is
T=0; and the length of the chord of contact is 2 2
2lr
l r
where ‘ r ’ is the radius of circle and ‘ l ’ is the
length of tangent.
4. Line and a Circle:
(i) If the distance of centre of circle 2 2 2 2 0x y gx fy c with radius r from the line 0lx my n
is p then
p r the line and circle have no common point
p r the line touches the circle. This is condition of tangency..
p r the line intersects the circle at 2 distinct points, i.e. it is a secant.
0p the line is a diameter..
the intercept made by circle on this line = 2 22 r p
and the angle subtended by the chord on the centre of the circle = 12cos /p r
(ii) Line y mx c touches the circle 2 2 2x y a if 2 2 21c a m
Thus, the tangent having slope m to the circle 2 2 2x h y k a is
2( ) 1y k m x h a m whose point of contact is 2 2,
1 1
am ah k
m m
This is called as Slope form of tangent.
(iii) Normal at the point (x1, y
1) on the circle S =0 is a line passing through its centre.
r
p
r (x ,y )11
1S
25E
K GYAN
M
AN D I R
mv
6. System of two Circles (with centres, C1 and C
2 ,distance ‘d’ apart, and radius r
1 and r
2 )
(i) do not touch each other (nor lie inside) if d > r1 + r
2 ; they have 4 common tangents.
Direct common tangents are intersecting at P1 and transverse common tangents are intersecting at
P2, as shown in the figure.
1 1 1 2 1
1 2 2 2 2
PC r PC
PC r PC
2r2C
1r
1CP1
P2
d
(ii) touch each other externally if d = r1 + r
2 ; and have 3 common tangents.
(iii) intersect at two distinct point if |r1 - r
2| < d < r
1 + r
2 and number common tangent is 2.
For the angle of intersection:
2 2 21 2
1 2
cos2
r r d
r r
Two circles 2 2 2 2 0i i ix y g x f y c , (i = 1,2) cut orthogonally i.e. 090
if 2g1g
2 + 2f
1f2 = c
1 + c
2
(iv) touch each other internally if d = |r1 - r
2| ; and have only 1 common tangent.
(v) one lies inside other if d < |r1 - r
2|.
7. Radical Axis: It is the locus of a point which moves so that the length of tangents drawn from it to two circlesS = 0, and S’ = 0, are equal, is given by S-S’ = 0.
This is same equation as that of the common chord if the circles are intersecting at two points, or that of thecommon tangent if the circles are touching.
8. Family of the circles:
(i) Equation of any circle passing through points of intersection of the circle S1 = 0 and S
2=0 is
1 2 0S S , where 1
(ii) Equation of any circle passing through points of intersection of the circle S = 0 and the line L = 0 is
0S L
9. Other Useful Concepts:
(i) Director circle: The locus of a point from which perpendicular tangents are drawn to the circle x2+y2
= r2 is x2+y2 = 2r2.
(ii) Pole and Polar: The equation of Polar of pole P(x1,y
1) w.r.t. a circle S = 0 is T = 0. Polar of Pole P
w.r.t. a Circle is the locus of points of intersection of tangents which are drawn at the points where avariable line passing through the Pole meets the Circle.
(iii) Two lines 1 1 1 0a x b y c , and 2 2 2 0a x b y c cut the coordinate axes in concyclic points if
1 2 1 2a a b b .
26E
K GYAN
M
AN D I R
mv
1. (b) Let the circle be 2 2 2( ) ( )x h y k r ....(i) which touches the circles externally
2 2 2x y a ....(ii), 2 2 2( 2 ) (2 )x a y a ....(iii) then
2 2h k r a ......(iv) 2 2( 2 ) 2h a k r a ....(v)
By (v) - (iv) 2 2 2 22h a k h k a 2 2 2 24( ) 9 24 16h k a ah h
Hence the locus of centre is 2 2 212 4 24 9 0x y ax a or 2 2 212( ) 4 3x a y a
2. (a) Equation of a circle passes through (1,0),(0,1) and (0, 0) is 2 2 0x y x y .
It also passes through (2k, 3k) then 2 513 5 0 ,0
13k k k i.e. for two values of k.
3. (c)r1 r2
60° 90°
As figure , 21
33 1
2 2
rAB BC r
1 23 2 3 2 2 2r r
then the radii of c1 and c
2 are 2 & 2 .
4. (b)
b
k
b
a a
2 2h b+( , )h k
2 2k a+
O
h
X
Y
As figure 2 2 2 2h b k a 2 2 2 2h k a b
then the equation of the locus of the centre of the circle
passing through the extremities of the two rods is x² y² = a² b².
8. (c) Let equation of circle 2 2 2 2 0S x y gx fy c . Now common chord of circle 0S &
2 2 4 0x y is 2 2 4 0gx fy c which will pass through (0,0) since 0S bisect the
circumferences of the circle 2 2 4 0x y then 4 0 4c c . Also common chord of circle 0S
& 2 2 2 6 1 0x y x y is (2 2) (2 6) 1 0g x f y c which will pass through (1, 3) since
0S bisect the circumferences of the circle then (2 2)1 3(2 6) 4 1 0 2 6 15 0g f g f .
The locus of the centre of the circle which bisects the circumferences of the circles is
2 6 15 0x y i.e. a straight line.
27E
K GYAN
M
AN D I R
mv
9. (b) 2 2(15 ) 30BOD BAO .Now
3cos30 , sin30
2 2
a aDO a BD a
3
1
2
2
a
a
15° A
B
C
30°o
Dthen coordinates of B & C are
3,
2 2
a a
and 3
,2 2
a a
10. (a) Distance between centres is 1 24 r r . Hence they touch
and common tangent by 1 2 0S S is 0x i.e. y axis.
(-3,0)
B
A (3,0)
C ( )0, 3
0x =
(1,0)
(0, 3)
1C
-
Other two direct common tangents intersect at
3(1) 1( 3) 3(0) 1(0)
, , 3,03 1 3 1
x y x y
and its equations are found as 3 3 0x y
and 3 3 0x y . These two tangents meet the
common tanget 0x at (0, 3)Q and (0, 3)R .Clearly ABC is equilateral .
Hence centroid of PQR is (1, 0).
28E
K GYAN
M
AN D I R
mv
PARABOLA
Locus of a point P which moves such that its distance from a fixed point S (called focus) bears a constant
ratio ‘e’ (called eccentricity) to its distance from a fixed line (called directrix), and where 1e
PS = PD, where D is the foot of perpendicular from P to the fixed line.
1. Standard equation of a Parabola:2 4y ax
(i) Focus, S = (a, 0), Vertex is (0, 0), Equation of axis 0y , Equation of tangent at vertex 0x , Equation
of directrix x a .
(ii) Focal chord of a parabola : Any chord passing through focus of the parabola. Latus rectum : A chordpassing through focus perpendicular to the axis of parabola. Its equation is x = a. Length of latus rectum =
4a = 4x distance between focus and vertex. End points of latus rectum are , 2L a a and ' , 2L a a
(iii) Double ordinate : A chord which is perpendicular to the axis of parabola.
(iv) Focal distance of any point 1 1( , )P x y : 1| |PS x a
K
2. Point 1 1,x y will be outside, on or inside the parabola depending on whether 21 14y ax is positive, zero or
negative.
3. Parametric equation of 2 4y ax 2x at 2y at
Equation of tangent at a point (at2, 2at) 2ty x at
and Equation of normal: 32y xt at at
4. Equation of tangent at a point (x1, y
1): 1 12yy a x x i.e. 0T
Equation of tangent in terms of slope m:a
y mxm
where the point of contact (a/m2, 2a/m)
29E
K GYAN
M
AN D I R
mv
5. Equation of normal at 1 1,P x y 11 1
2
yy y x x
a
and equation of normal in terms of slope m: 32y mx am am where the foot of normal is (am2, - 2am)
6. Properties of normalsNumber of normal : y = mx - 2am - am3 is a cubic equation in m. So in general at most three normals canpass through a point.The sum of slopes of these normals is zero and also sum of the ordinates of the feet of normal is zero.
7. Properties of focal chords:
Tangents at the extremities of any focal chord intersect at right angles on the directrix 1 2 1t t
8. Other Parabolas: To solve problems related to parabolas other than 2 4y ax , we have to use conceptss
like shifting of origin etc, e.g. 2
4y a x is the equation of that parabola whose vertex
is , and whose axis is parallel to x-axis.
2
4x a y is the equation of that parabola whose vertex is , and whose axis is parallel to
y-axis.
9 . Other important points:
Point of intersection of tangents at t1 and t
2 is 1 2 1 2, ( )at t a t t . .i e (GM, AM)
Point of intersection of normal at t1 and t
2 is 2 2
1 2 1 2 1 2 1 2{2 ( ), ( )}a a t t t t at t t t
This point will be the parabola if 1 2 2t t
If the normal at the point (at1
2, 2at1) meets the parabola again in the point (at
22, 2at
2), then 2 1 1(2 / )t t t
Equation of pair of tangents of a parabola drawn from a given point (x1, y
1) is 2
1T SS , Chord of contact
0T , Polar 0T
Chord of parabola whose mid point is (x1, y
1) is 1T S
PD SP ST SN , DPT SPT , 090PSR PKS
30E
K GYAN
M
AN D I R
mv
1. (d) Length of semilatus rectum is H.m. of two segments of a focal chord then 2(3)(5) 15
23 5 8
a a
2. (c) Coordiinate of P & Q are 21 1,2at at and 2
2 2( ,2 )at at
Now slope of OP is
1
2
t and slope of OQ is
2
2
t then
1 2
2 2. 1
t t
P
R
Q
O
1 2 4t t ....(i)
Now equation of line PQ is 21 2 1 1 2 12 ( ) ( )2x t t y at t t at
1 22at t
Now by (i) 1 22 ( ) 8x t t y a . So coordinate of point R is (4 ,0)a i.e. 4OR a .
3. (c) Any tangent to 2 4( 1), 1y x a is1
( 1)y m xm
........(i)
Any tangent to 2 8( 2); 2y x a is 2
( 2)y M xM
but tangents to the parabolas are perpendicular to each other then 1 1/Mm M m
i.e.1
( 2) 2y x mm
........(ii)
On subtracting we get 1 1
0 3 0m x mm m
or 3 0x
4. (a) Equationof circle taking OP as diameter is
2 2 2 21 1 1 12 0 2 0x x at y y at x y at x at y
Similarly equation of circle taking OQ as diameter is
R
1
2
21 1( ,2 )at at
22 2( , 2 )at at
2 2 22 22 0x y at x at y
The equation of OR is 2 21 2 1 22 0a t t x a t t y
1 21 2 1 2( ) 2 0 tan 2 tan
2
t tt t x y t t
..........(i)
Now equation of tangent at point 21 1( ,2 )at at is 2
1 1yt x at
i.e. 1 1
1
1tan cot t
t . Similarly 2 2cot t .By (i) we get 1 2cot cot 2 tan
5.(a) If one end of focal chord of parabola 2 4y x is 2 ,2t t then coordinates of ther end is 2
1 2,
t t
then
length of focual chord is
2 2 2 2 4 2
2
2
1 2 1 1 1 12 4t t t t t t
t t t t t t
31E
K GYAN
M
AN D I R
mv
6. (d) One end of normal chord of parabola 2 4y ax is 21 1( , 2 )P at at then other end is
2
1 1
1 1
2 2,2Q a t a t
t t
.Now slope of
1
2OP
t then slope of OQ is
1
1
2
2t
t
Since the normal chord at a point ' t1' subtends a right angle at the vertex therefore
2 21 1
11
1
2 2. 1 4 2 2
2t t
t tt
8. (d) Slope of 2
2
1
tSP
t
,slope of
2
2
1
tSQ
t
since 2
PSQ
. Now
2
21
tan 45 2 12
a tt t
at
S M(a,0)
2
2
(
(
,
,
2
0
)
)
a
a
t
t
at
2( , 2 )at at-Q
P45°
45°2 2 2 2
2 1 0 2 12
t t t
tangent at P is 2yt x at , tangent at Q is 2yt x at
Now intersection pont of tangents is 22 ,0 2 1 ,0at a
9. (c) Normal at point 22 ,4t t cut parabola again at
22 2
2 ,4t tt t
then
24 4
22
t tt
21 2t
t then ends of normal are 8, 8 and 18,12 then slope of the normal chord is 2.
10. (a) In 2
,2
aMDS
at
02 1, tan30 32 3
aMDS t
at
M P
S(a,0)D
( ,2 )a at-2( ,2 )at at
30°
then coordinate of P is 3 ,2 3a a
Now 22
3 2 3 4SP a a a a