Notes For Bachelor of Science (TU)
BSC NOTES
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Lecture Note
By: Chet Nath Tiwari
Trichandra Multiple Campus
For: B Sc. 1st year
Two Dimentional, Analytical Geometry
Topics covered: Transformation, Rotations, Process involving combination of
translation of axes, Invariants in orthogonal transfromation
1
TRANSFORMATION OF COORDINATES
It is sometimes found desirable in the discussion of problems to alter the origin and axes of
coordinates, either by altering the origin without alteration of the direction of axes or by
alteration of the axes and keeping the origin unchanged or both. Either of these process is
called a transformation of coordinates.
1.1
To change the origin of coordinates without changing the directions of the axes
Let Ox, Oy be the orininal rectangular axes and let O0 X and O0 Y be new axes, parallel to
1
the original axes. Let the coodinates of O0 be (h, k).
Let P be any point in the plane. Let (x, y) and (X, Y ) be the co-ordinates of P with referred
to orinal and new axes respectively. Draw O0 Q and P R perpendiculars on Ox such that P R
meets O0 X at T . Then
x = OR = OQ + QR = OQ + O0 T = h + X
y = RP = RT + T P = QO0 +0 T P = k + Y
Hence,
(old x-coordinate) = (new x-coordinate) + h
and
(old y-coordinate) = (new y-coordinate) + k
If therefore the origin is shifted at a point (h, k) we must substitute X + h and Y + k for x and
y respectively.
The transformation formula from new axes to old axes is
X = x − h,
Y = y − k.
Example 1 Transform to parallel axes through the point (−1, 2) the equation x2 −y 2 +2x+4y =
0.
Solution: The given equation is
x2 − y 2 + 2x + 4y = 0
(1)
Let (h, k) = (−1, 2) be the new origin. Then substituting x = X + h = X − 1 and y = Y + k =
Y + 2 in (1).
(X − 1)2 − (Y + 2)2 + 2(X − 1) + 4(Y + 2) = 0
or, X 2 − 2X + 1 − (Y 2 + 4Y + 4) + 2X − 2 + 4Y + 8 = 0
or, X 2 − Y 2 + 3 = 0
Example 2 Find the equation of the curve 9x2 + 4y 2 + 18x − 16y = 11 referred to parallel axes
through (−1, 2).
Solution: The given equation is
9x2 + 4y 2 + 18x − 16y − 11 = 0
(2)
Let (h, k) = (−1, 2) be the new origin. Then substituting x = X + h = X − 1 and y = Y + k =
Y + 2 in (2).
9(X − 1)2 + 4(Y + 2)2 + 18(X − 1) − 16(Y + 2) − 11 = 0
or, 9X 2 − 18X + 9 + 4Y 2 + 16Y + 16 + 18X − 18 − 16Y − 32 − 11 = 0
or, 9X 2 + 4Y 2 − 36 = 0
Which is required equation.
2
Example 3 Transform the equation 2x2 + 4xy + 5y 2 − 4x − 22y + 7 = 0 to parallel axes through
(−2, 3).
Solution: The given equation is
2x2 + 4xy + 5y 2 − 4x − 22y + 7 = 0
(3)
Let (h, k) = (−2, 3) be the new origin. Then substituting x = X + h = X − 2 and y = Y + k =
Y + 3 in (3).
2(X − 2)2 + 4(X − 2)(Y + 3) + 5(Y + 3)2 − 4(X − 1) − 22(Y + 3) + 7 = 0
or, 2X 2 − 8X + 8 + 4XY − 8Y + 12X − 24 + 5Y 2 + 30Y + 45 − 4X + 8 − 22Y − 66 + 7 = 0
or, 2X 2 + 4XY + 5Y 2 − 22 = 0
Which is required equation.
Example 4 What does the equation 2x2 + y 2 − 4x + 4y = 0 becomes when it is transferred to
parallel axes through the point (1, −2)?
Solution: The given equation is
2x2 + y 2 − 4x + 4y = 0
(4)
Let (h, k) = (1, −2) be the new origin. Then substituting x = X + h = X + 1 and y = Y + k =
Y − 2 in (4).
2(X + 1)2 + (Y − 2)2 − 4(X − 1) + 4(Y − 2) = 0
or, 2X 2 + 4X + 2 + Y 2 − 4Y + 4 − 4X − 4 + 4Y − 8 = 0
or, 2X 2 + Y 2 − 6 = 0
Which is required equation.
Example 5 What does the equation (x − a)2 + (y − b)2 = c2 becomes when it is transferred to
parallel axes through the point (a − c, b)?
Solution: The given equation is
(x − a)2 + (y − b)2 = c2
(5)
Let (h, k) = (a − c, b) be the new origin. Then substituting x = X + h = X + a − c and
y = Y + k = Y + b in (6).
(X + a − c − a)2 + (Y + b − b)2 = c2
or, (X − c)2 + Y 2 = c2
or, X 2 − 2cX + c2 + Y 2 = c2
or, X 2 − 2cX + Y 2 = 0
Which is required equation.
3
Example 6 What does the equation (x − a)2 + (y − b)2 = c2 becomes when it is transferred to
parallel axes through the point (a, b − c)?
Solution: The given equation is
(x − a)2 + (y − b)2 = c2
(6)
Let (h, k) = (a, b − c) be the new origin. Then substituting x = X + h = X + a and y = Y + k =
Y + b − c in (4).
(X + a − a)2 + (Y + b − c − b)2 = c2
or, X 2 + (Y − c)2 = c2
or, X 2 + Y 2 − 2cY + c2 = c2
or, X 2 + Y 2 − 2cY = 0
Which is required equation.
Example 7 What does the equation (a − b)(x2 + y 2 ) − 2abx = 0 becomes if the origin be moved
ab
to the point a−b
,0 ?
Solution: The given equation is
(a − b)(x2 + y 2 ) − 2abx = 0
ab
Let (h, k) = a−b
, 0 be the new origin. Then substituting x = X + h = X +
y = Y + k = Y in (7).
#
"
2
ab
ab
2
+ Y − 2ab X +
=0
(a − b) X +
a−b
a−b
(7)
ab
a+b
and
ab
a2 b 2
2a2 b2
2
or, (a − b) X + 2X
+
+
Y
−
2abX
−
=0
a − b (a − b)2
a−b
2
or, (a − b)X 2 + 2abX +
a2 b 2
2a2 b2
+ (a − b)Y 2 − 2abX −
=0
a−b
a−b
or, (a − b)X 2 + (a − b)Y 2 −
a2 b 2
=0
a−b
or, (a − b)2 X 2 + (a − b)2 Y 2 − a2 b2 = 0
Which is required equation.
Example 8 The equation of a curve referred to new axes, axes retaining their directions, and
origin (−1, 2) is 2X 2 + Y 2 − 6 = 0. Find the equation referred to the original axes.
4
Solution: The given transformed equation is
2X 2 + Y 2 − 6 = 0
(8)
Let (h, k) = (1, −2) be new origin. Then
y =Y +k =Y −2
x = X + h = X + 1,
X = x − 1,
Y =y+2
Putting the values of X and Y in (8)
2(x − 1)2 + (y + 2)2 − 6 = 0
or, 2(x2 − 2x + 1) + y 2 + 4y + 4 − 6 = 0
or, 2x2 + y 2 − 4x + 4y = 0
which is required equation.
1.2
To change the origin without any change in the direction of the
axes so that the equation ax2 + 2hxy + by 2 + 2gx + 2f y + c = 0
transform in to one, in which there is no first degree terms
The general second degree equation in x, y is
ax2 + 2hxy + by 2 + 2gx + 2f y + c = 0
(9)
Let (α, β) be new origin. Then
x = X + α,
Y =y+β
Putting the values of x and y in (9) we have
a(X + α)2 + 2h(X + α)(Y + β) + b(Y + β)2 + 2g(X + α) + 2f (Y + β) + c = 0
or, aX 2 + 2hXY + bY 2 + 2(aα + hβ + g)X + 2(hα + bβ + f )Y +
(aα2 + 2hαβ + bβ 2 + 2gα + 2gβ + c) = 0
The transformed equation will be free from first degree term if
aα + hβ + g = 0
(10)
hα + bβ + f = 0
(11)
5
Solving (9) and 10 by method of cross-multiplication
α
β
1
=
=
hf − bg
gh − af
ab − h2
Then
hf − bg
ab − h2
Thus to remove the first degree term of
β=
α=
gh − af
ab − h2
ax2 + 2hxy + by 2 + 2gx + 2f y + c = 0
we have to shifting the origin by
(α, β) =
hf − bg gh − af
,
ab − h2 ab − h2
where ab − h2 6= 0.
Example 9 By transforming to parallel axes through a properly chosen point (h, k) , prove that
the equation
12x2 − 10xy + 2y 2 + 11x − 5y + 2 = 0
can be reduced to one containing only terms of second degree.
Solution: The given equation is
12x2 − 10xy + 2y 2 + 11x − 5y + 2 = 0
Let (h, k) be the new origin. Then
x = X + h,
y =Y +k
Putting the values of x, y in the given equation, the transformed equation is
12(X + h)2 − 10(X + h)(Y + k) + 2(Y + k)2 + 11(X + h) − 5(Y + k) + 2 = 0
or, 12X 2 − 10XY + 2Y 2 + (24h − 10k + 11)X + (−10h + 4k − 5)Y +
12h2 − 10hk + 2k 2 + 11h − 5k + 2 = 0
To remove the first degree term, we must have, coefficients of X and Y must be zero
24h − 10k + 11 = 0
−10h + 4k − 5 = 0
6
Solving these two equations by method of cross-multiplication, we get
h
k
1
=
=
50 − 44
−110 + 120
96 − 100
∴
3
h=− ,
2
k=−
5
2
For this value of h and k, we have
2
9
3
5
5
12h − 10hk + 2k + 11h − 5k + 2 = 12
− 10 −
−
+2 −
4
2
2
2
3
−5
+11 −
−5
+2
2
2
2
2
54 75 25 33 25
−
+
−
+
+2=0
2
2
2
2
2
Hence, the given equation reduces to 12X 2 − 10XY + 2Y 2 = 0, whern origin is shifted to
−3 −5
,
.
2
2
=
Example 10 Translate the axes so as to change the equation 3x2 −2xy +4y 2 +8x−10y +8 = 0
into an equation with linear terms missing.
Solution: The given equation is
3x2 − 2xy + 4y 2 + 8x − 10y + 8 = 0
Let (h, k) be the new origin. Then
x = X + h,
y =Y +k
Putting the values of x, y in the given equation, the transformed equation is
3(X + h)2 − 2(X + h)(Y + k) + 4(Y + k)2 + 8(X + h) − 10(Y + k) + 8 = 0
or, 3X 2 − 2XY + 4Y 2 + (6h − 2k + 8)X + (−2h + 8k − 10)Y +
3h2 − 2hk + 4k 2 + 8h − 10k + 8 = 0
For linear term missing, we must have, coefficients of X and Y must be zero
6h − 2k + 8 = 0
−2h + 8k − 10 = 0
7
(12)
Solving these two equations by method of cross-multiplication, we get
h
k
1
=
=
20 − 64
−16 + 60
48 − 4
∴
h = −1,
k=1
For this value of h and k, we have
3h2 − 2hk + 4k 2 + 8h − 10k + 8 = 3 + 2 + 4 − 8 − 10 + 8 = −1
Hence, the given equation (12) reduces to 3X 2 − 2XY + 4Y 2 − 1 = 0, whern origin is shifted
to (−1, 1).
8
1.3
To change the direction of the axes without changing the origin.
Let Ox and Oy be the original sytem of axes and OX and OY be the new system of axes..
Let ∠xoX = θ, through which the axes are turned. Let P be any point on the plane. Let (x, y)
and (X, Y ) be the the coordinates of the point P referred to the original axes and new axis
OX, OY repectively. Draw P S and P R perpendicular to ox and Ox. Also draw RT and RQ
perpendiculars to Ox and P S. From figure x = Os, y = P s and OR = X, P R = Y
∠QP R =
π
− ∠P RQ = ∠ROT = θ
2
Now
x = OS = OT − QR = OR cos θ − P R sin θ = X cos θ − Y sinθ
and
y = P S = P Q + QS = P Q + RT
= P R cos θ + OR sin θ
= X sin θ + Y cos θ
Hence the required transformation when the axes are turn through an angle θ without change
of origin are
x = X cos θ − Y sinθ
y = X sin θ + Y cos θ
This transformation is known as a rotation.
Example 11 Transform the equation x2 +2cxy +y 2 = a2 by turnining rectangular axes through
the angle π4 .
9
Solution: The given equation is
x2 + 2cxy + y 2 = a2
(13)
Since the axes are turned by an angle θ = π4 . Hence
x = X cos θ − Y sin θ = X cos
π
1
π
− Y sin = √ (X − Y )
4
4
2
π
1
π
+ Y cos = √ (X + Y )
4
4
2
Putting the values of x and y in the equation (13),
y = X sin θ + Y cos θ = X sin
(X − Y )(X + Y ) 1
1
(X − Y )2 + 2c
+ (X + Y )2 = a2
2
2
2
or, (1 + c)X 2 + (1 − c)Y 2 = a2
which is required equation.
√
Example 12 What does the equation x2 + 2 3xy − y 2 = 2a2 becomes when the axes are turned
through an angle of 300 to the original axes.
Solution: The given equation is
√
x2 + 2 3xy − y 2 = 2a2
Since the axes are turned by an angle θ = 300 . Hence
1 √
x = X cos θ − Y sin θ = X cos 300 − Y sin 300 = ( 3X − Y )
2
√
1
y = X sin θ + Y cos θ = X sin 300 + Y cos 300 = (X + 3Y )
2
Putting the values of x and y in the equation (14),
" √
#
√
√
√
1 √
(
3X
−
Y
)
(X
+
3Y
)
1
( 3X − Y )2 + 2 3
− (X + 3Y )2 = a2
4
2
2
4
√
i 1
√
√
1 √
2 3h √
2
or, ( 3X − Y ) +
( 3X − Y )(X + 3Y ) − (X + 3Y )2 = 2a2
4
4
4
√
√
√
√
or, 3X 2 − 2 3XY + Y 2 + 2 3( 3X 2 + 3XY − XY − 3Y 2 )
√
−(X 2 + 2 3XY + 3Y 2 ) = 8a2
or, X 2 − Y 2 = a2
which is required equation.
10
(14)
Example 13 What does the equation of the straight lines 7x2 + 4xy + 4y 2 = 0 becomes when
the axes are turned through 450 , the origin remain fixed ?
Solution: The given equation is
7x2 + 4xy + 4y 2 = 0
(15)
Since the axes are turned by an angle θ = 450 . Hence
1
x = X cos θ − Y sin θ = X cos 450 − Y sin 450 = √ (X − Y )
2
1
y = X sin θ + Y cos θ = X sin 450 + Y cos 450 = √ (X + Y )
2
Putting the values of x and y in the equation (15),
1
(X − Y ) (X + Y )
1
2
√
√
7 (X − Y ) + 4
+ 4 (X + Y )2 = 0
2
2
2
2
or, 7(X 2 − 2XY + Y 2 ) + 4(X 2 − Y 2 ) + 4(X 2 + 2XY + Y 2 ) = 0
or, 15X 2 − 6XY + 7Y 2 = 0
which is required equation.
Example 14 If the axes are turned through an angle tan−1 1. what does the equation 4xy −
3x2 = a2 becomes?
Solution: The given equation is
4xy − 3x2 = a2
Since the axes are turned by an angle θ = tan−1 1 = 450 . Hence
1
x = X cos θ − Y sin θ = X cos 450 − Y sin 450 = √ (X − Y )
2
1
y = X sin θ + Y cos θ = X sin 450 + Y cos 450 = √ (X + Y )
2
Putting the values of x and y in the equation (16),
4·
X −Y X +Y
(X − Y )2
√
√
−3
= a2
2
2
2
or, 4(X 2 − Y 2 ) − 3(X − Y )2 = 2a2
or, X 2 + 6XY − 7Y 2 = 2a2
which is required equation.
11
(16)
Example 15 If the axes are turned through an angle α. what does the equation x2 +2xy tan2 α−
y 2 = a2 become?
Solution: The given equation is
x2 + 2xy tan2 α − y 2 = a2
(17)
Since the axes are turned by an angle θ = α. Hence
x = X cos θ − Y sin θ = X cos α − Y sin α
y = X sin θ + Y cos θ = X sin α + Y cos α
Putting the values of x and y in the equation (17),
(X cos α − Y sin α)2 + 2(X cos α − Y sin α)(X sin α + Y cos α) tan2 α
−(X sin α + Y cos α)2 = a2
or, X 2 cos2 α − 2XY cos α sin α + Y 2 sin2 α + 2 tan2 α(X 2 sin α cos α + XY cos2 α − XY sin2 α)
−2Y 2 tan2 α sin α cos α − X 2 sin2 α − 2XY sin α cos α − Y 2 cos2 α = a2
or, X 2 (cos2 α − sin2 α) − Y 2 (cos2 α − sin2 α) − 4XY sin α cos α+
2 tan2 α[XY (cos2 α − sin2 α) + (X 2 − Y 2 ) sin α cos α] = a2
or,(X 2 − Y 2 ) cos 2α − 2XY sin 2α + tan2 α 2XY cos 2α + (X 2 − Y 2 ) sin 2α = a2
which is required equation.
Example 16 Transform the follwing equation to the axes inclined at 450 to the original axes
y 4 + x4 + 6x2 y 2 = 2 [BA/B.Sc.2064].
Solution: The given equation is
y 4 + x4 + 6x2 y 2 = 2
(18)
Let (X, Y ) be the coordinates of the point referredto new axes, which is rotated by θ = 450 .
Then
X −Y
x = X cos θ − Y sin θ = X cos 450 − Y sin 450 = √
2
X +Y
y = X sin θ + Y cos θ = X sin 450 + Y cos 450 = √
2
The equation (18)
(X + Y )4 (X − Y )4
(X − Y )2 (X + Y )2
+
+6
=2
4
4
2
2
12
or, (X 2 + Y 2 + 2XY )2 + (X 2 + Y 2 − 2XY )2 + 6(X 2 − Y 2 )2 = 8
(19)
Using (a + b)2 + (a − b)2 = 2(a2 + b2 ) we get
(X 2 + Y 2 + 2XY )2 + (X 2 + Y 2 − 2XY )2 = 2[(X 2 + Y 2 )2 + (2XY )2 ]
From (19),
2 (X 2 + Y 2 )2 + (2XY )2 + 6(X 2 − Y 2 )2 = 8
or, X 4 + 2X 2 Y 2 + Y 4 + 4X 2 Y 2 + 3(X 4 − 2X 2 Y 2 + Y 4 ) = 4
or, 4(X 4 + Y 4 ) = 4
or, X 4 + Y 4 = 1
which is required equation.
Example 17 If the axes be turned through an angle tan−1 2, what does the equation 4xy−3x2 =
a2 becomes?
Solution: The given equation is
4xy − 3x2 = a2
Let the angle of rotation be θ = tan−1 2. Then tan θ = 2.
1
1
1
1
=√
=√
=√
2
2
sec θ
1+2
5
1 + tan θ
r
√
1
2
sin θ = 1 − cos2 θ = 1 − = √
5
5
−1
Since the axes are turned by an angle θ = tan 2. Hence
cos θ =
1
x = X cos θ − Y sin θ = √ (X − 2Y )
5
1
y = X sin θ + Y cos θ == √ (2X + Y )
5
Putting the values of x and y in the equation (20),
1
1
(X − 2Y )2
√
√
4
(X − 2Y )
(2X + Y ) − 3
= a2
5
5
5
or, 4(X − 2Y )(2X + Y ) − 3(X − 2Y )2 = 5a2
or, 8X 2 − 12XY − 8Y 2 − 3X 2 + 12XY − 12Y 2 = 5a2
or, 5X 2 − 20Y 2 = 5a2
or, X 2 − 4Y 2 = a2
which is required equation.
13
(20)
Example 18 If the axes be turned through an angle tan−1 3, what does the equation 3xy−4y 2 =
a2 becomes?
Solution: The given equation is
3xy − 4y 2 = a2
(21)
Let the angle of rotation be θ = tan−1 3. Then tan θ = 3.
1
1
1
1
√
√
=√
=
=
sec θ
1 + 32
10
1 + tan2 θ
r
√
1
3
2
sin θ = 1 − cos θ = 1 −
=√
10
10
cos θ =
Since the axes are turned by an angle θ = tan−1 3. Hence
1
x = X cos θ − Y sin θ = √ (X − 3Y )
10
1
y = X sin θ + Y cos θ == √ (3X + Y )
10
Putting the values of x and y in the equation (20),
1
1
(3X + Y )2
= a2
3 √ (X − 3Y ) √ (3X + Y ) − 4
10
10
10
or, 3(X − 3Y )(3X + Y ) − 4(3X + Y )2 = 10a2
or, 9X 2 − 24XY − 9Y 2 − 36X 2 − 24XY − 4Y 2 = 10a2
or, − 27X 2 − 48XY − 13Y 2 = 10a2
or, 27X 2 + 48XY + 13Y 3 + 10a2 = 0.
which is required equation.
Example 19 Find The angle through which the axes may turned so that equation Ax+By+C =
0 may reduced to the form X = constant, and determine the value of this constant.
Solution:The given equation is
Ax + By + C = 0
Let θ be the angle of rotation of the axes. Then
x = X cos θ − Y sin θ
14
(22)
y = X sin θ + Y cos θ
Putting values of x, y in (22)
A(X cos θ − Y sin θ) + B(X sin θ + Y cos θ) + C = 0
or, X(A cos θ + B sin θ) + Y (B cos θ − A sin θ) + C = 0
The equation (23) becomes X = constant, if coefficient of Y = 0. i.e.
B cos θ − A sin θ = 0
B
=⇒ θ = tan−1
or, tan θ =
A
cos θ =
1
1
1
q
=√
=
sec θ
1 + tan2 θ
1+
B
A
A
=√
A2 + B 2
B 2
A
A
B2
sin θ = 1 − cos θ = 1 − 2
= 2
A + B2
A + B2
B
∴, sin θ = √
2
A + B2
2
2
2
Putting the values of sin θ and cos θ in (23)
B
A
+ B√
+Y ·0+C =0
X A√
A2 + B 2
A2 + B 2
A2 + B 2
or, X √
= −C
A2 + B 2
−C
or, X = √
A2 + B 2
Which is in the form of X = k where k =
√ −C
A2 +B 2
15
(23)
1.4
To change the direction of axes along with change of origin
(Combination of Translation and Rotation)
Let the origin O be transfer to the point O0 (h, k) without changing the direction of the axes.
Let P be any point in the plane and let its coordinates be (x, y) and (X 0 , Y 0 ) referred to the
original and new axes respectively. Then
x = X 0 + h,
y =Y0+k
(24)
Let the new axes rotate about O0 through an angle θ, and let the coordinates of P be (X, Y )
.Then
X 0 = X cos θ − Y sin θ
Y 0 = X sin θ + Y cos θ
Substituting the values of X 0 and Y 0 in 24, we have
x = X cos θ − Y sin θ + h
y = X sin θ + Y cos θ + k
Thus if the origin be transferred to (h, k) and the axes turned through an angle θ, the transformation will be given by
x = X cos θ − Y sin θ + h
y = X sin θ + Y cos θ + k
16
Example 20 What does the equation 2x2 + 4xy − 5y 2 + 20x − 22y − 14 = 0 becomes when
referred to rectangular axes through the point (−2, −3), the new axes being inclined at an angle
π
, with the old?
4
Solution: The given equation is
2x2 + 4xy − 5y 2 + 20x − 22y − 14 = 0
(25)
Let (h, k) = (−2, −3) be new origin. Then
x = X 0 + h = X 0 − 2, and y = Y 0 + k = Y 0 − 3
Putting the values ofx and y in (25)
2(X 0 − 2)2 + 4(X 0 − 2)(Y 0 − 3) − 5(Y 0 − 3)2 + 20(X 0 − 2) − 22(Y 0 − 3) − 14 = 0
or, 2X 02 − 4X 0 Y 0 − 5Y 02 − 8X 0 − 12X 0 + 20X 0 − 8Y 0 + 30Y 0 − 22Y 0
+8 + 24 − 45 − 40 + 66 − 14 = 0
or, 2X 02 − 4X 0 Y 0 − 5Y 02 − 1 = 0
Again,θ =
π
4
(26)
is the angle of rotation,
X 0 = X cos θ − Y sin θ = X cos
π
X −Y
π
− Y sin = √
4
4
2
Y 0 = X cos θ + Y sin θ = X cos
π
X +Y
π
+ Y sin = √
4
4
2
Putting the values of X 0 and Y 0 in (26)
2
X −Y X +Y
(X − Y )2
(X + Y )2
√
+4 √
−5
−1=0
2
2
2
2
or, 2(X 2 − 2XY + Y 2 ) + 4(X 2 − Y 2 ) − 5(X 2 + 2XY + Y 2 ) − 2 = 0
or, X 2 − 14XY − 7Y 2 − 2 = 0
Example 21 Transform the equation x2 − 2xy + y 2 + x − 3y = 0 to axes through the point
(−1, 0) parallel to the lines bisecting the angle between the original axes.
Solution: The given equation is
x2 − 2xy + y 2 + x − 3y = 0
Let (h, k) = (−1, 0) be new origin. Then
x = X 0 + h = X 0 − 1, and y = Y 0 + 0 = Y 0
17
(27)
Putting the values ofx and y in (27)
(X 0 − 2)2 − 2(X 0 − 2)(Y 0 − 3) + Y 02 + X 0 − 2 − 3Y 0 = 0
or, X 02 − 2X 0 Y 0 + Y 02 − 4X 0 + 6X 0 + X 0 + 4Y 0 − 3Y 0 = 0
or, X 02 − 2X 0 Y 0 + Y 02 − X 0 − Y 0 = 0
(28)
Again,θ = 450 is the angle of rotation,
X 0 = X cos θ − Y sin θ = X cos 450 − Y sin 450 =
X −Y
√
2
Y 0 = X cos θ + Y sin θ = X cos 450 + Y sin 450 =
X +Y
√
2
Putting the values of X 0 and Y 0 in (28)
X −Y X +Y
(X + Y )2 X − Y
X +Y
(X − Y )2
√
−2 √
− √
+
− √
=0
2
2
2
2
2
2
√
√
X 2 − 2XY + Y 2 − 2(X 2 − Y 2 ) + (X 2 + 2XY + Y 2 ) − 2(X − Y )) − 2(X + Y )
=0
or,
2
√
or, 4Y 2 − 2 2X = 0
√
or, X − 2Y 2 = 0
which is required equation.
1.5
Removal of xy term from ax2 + 2hxy + by 2 , the axes being rectangular
The homogeneous function of second degree in x and y is
ax2 + 2hxy + by 2
We shall assume the origin to be fixed and rotate the axes through an angle θ. Then
x = X cos θ − Y sin θ
y = X sin θ + Y cos θ
where X, Y are new coordinate axes.
Substituting values of x, y in (29)
ax2 + 2hxy + by 2
= a(X cos θ − Y sin θ)2 + 2h(X cos θ − Y sin θ)(X sin θ + Y cos θ)
18
(29)
+b(X sin θ + Y cos θ)2
= (a cos2 θ + 2h cos θ sin θ + b sin2 θ)X 2 + (a sin2 θ − 2h cos θ sin θ + b cos2 θ)Y 2 +
2[(b − a) cos θ sin θ + h(cos2 θ − sin2 θ)]
For the removal of XY term, we have
Coefficient of XY = 0
or, 2(b − a) cos θ sin θ + h(cos2 θ − sin2 θ) = 0
or, (a − b) sin 2θ = 2h cos θ
2h
tan 2θ =
a−b
Therefore, the required angle θ is given by
tan 2θ =
a−b
2h
or cot 2θ =
a−b
2h
Example 22 Find the angle through which the axes must be rotated to remove the term con√
taining xy in 3x2 + 2xy + 3y 2 − 2x = 0. Also, find the transformed equation.
Solution: The given equation is
3x2 + 2xy + 3y 2 −
√
2x = 0
(30)
Let θ be the angle between new and old axes. Then
x = X cos θ − Y sin θ
y = X sin θ + Y cos θ
Putting the values of x, y in the equation (30), we have
3(X cos θ − Y sin θ)2 + 2(X cos θ − Y sin θ)(X sin θ + Y cos θ) +
√
3(X sin θ + Y cos θ)2 − 2(X cos θ − Y sin θ) = 0
or, (3 cos2 θ + 2 sin θ cos θ + 3 sin2 θ)X 2 + (3 sin2 θ − 2 sin θ cos θ + 3 cos2 θ)Y 2
√
√
+(−6 sin θ cos θ + 2 cos2 θ − 2 sin2 θ + 6 sin θ cos θ)XY − 2 cos θX + 2 sin θY = 0
For the removal of XY term we have
−6 sin θ cos θ + 2 cos2 θ − 2 sin2 θ + 6 sin θ cos θ = 0
Or, cos2 θ − sin2 θ = 0
or, cos 2θ = 0
or, 2θ = 900
or, = θ = 450
19
(31)
Substituting the values of θ in equation (31))
1
1 1
1
1 1
1
1
2
3
+ 2√ √ + 3
X + 3
− 2√ √ + 3
Y2
2
2
2
2
2 2
2 2
√
√
1
1
=0
− 2 √ X+ 2 √
2
2
or, 4X 2 + 2Y 2 − X + Y = 0.
Example 23 By suitable translation and rotation, transform the equation 8x2 + 4xy + 5y 2 −
2
2
24x − 24y = 0 to the form xa2 + yb2 = 1.
Solution: The given equation is
8x2 + 4xy + 5y 2 − 24x − 24y = 0
(32)
First we remove the linear term:
Let us transform the origin to new (h, k). Then
x = X 0 + h,
y =Y0+k
Substituting the values of x, y in (32)
8(X 0 + h)2 + 4(X 0 + h)(Y 0 + k) + 5(Y 0 + k)2 − 24(X 0 + h) − 24(Y 0 + k) = 0
or, (8X 02 + 4X 0 Y 0 + 5Y 02 ) + (8h2 + 4hk + 5k 2 − 24h − 24k) +
(16h + 4k − 24)X 0 + (4h + 10k − 24)Y 0 = 0
(33)
To remove the linear terms we must have
16h + 4k − 24 = 0
4h + 10k − 24 = 0
i.e.
4h + k − 6 = 0
(34)
4h + 10k − 24 = 0
(35)
Solving (34)) and (35) we get h = 1 and k = 2 . Putting the values of h and k in (33), we get
(8X 02 + 4X 0 Y 0 + 5Y 02 ) + (8 · 12 + 4 · 1 · 2 + 5 · 22 − 24 · 1 − 24 · 2) + 0.X 0 + 0 · Y 0 = 0
or,8X 02 + 4X 0 Y 0 + 5Y 02 = 36
Second we remove X 0 Y 0 :
Let the axes be rotate by an angle θ. Then
X 0 = X cos θ − Y sin θ
20
(36)
Y 0 = X sin θ + Y cos θ
Putting these values in 36
8(X cos θ − Y sin θ)2 + 4(X cos θ − Y sin θ)(X sin θ + Y cos θ) + 5(X sin θ + Y cos θ)2 = 36
or,(8 cos2 θ + 4 sin θ cos θ + 5 sin2 θ)X 2 + (8 sin2 θ − 4 sin θ cos θ + 5 cos2 θ) +
(−16 cos θ sin θ + 4 cos2 θ − 4 sin2 θ + 10 sin θ cos θ)XY = 36
or, (8 cos2 θ + 2 sin 2θ + 5 sin2 θ)X 2 + (8 sin2 θ − 2 sin 2θ + 5 cos2 θ) +
(−3 sin 2θ + 4 cos 2θ)XY = 36
To remove XY term,
Coefficient of XY = 0
or, − 3 sin 2θ + 4 cos 2θ = 0
3
tan 2θ =
4
1
1
1
q
=
∴ cos 2θ =
=√
sec 2θ
1 + tan2 2θ
1+
p
cos 2θ = 1 − sin2 2θ =
r
1−
=
9
16
4
5
16
3
=
25
5
Also,
1+
1 + cos 2θ
=
2
2
3
5
4
5
4
1
sin2 θ = 1 − cos2 θ = 1 − =
5
5
2
2
Putting values of cos 2θ, sin 2θ, sin θ, cos θ in (37)
4
4
4
4
2
8 · + 2 · + 1 X + 8 · − 2 · + 1 Y 2 = 36
5
5
5
5
cos2 θ =
45 2 20 2
X + Y = 36
5
5
2
or, 9X + 4Y 2 = 36
or,
or,
which is in the form
X2 Y 2
+
=1
4
9
X2 Y 2
+ 2 =1
a2
b
where a2 = 4 and b2 = 9.
21
=
(37)
2
Invarients
Example 24 If any change of axes, without change of origin, ax2 + 2hxy + by 2 becomes AX 2 +
2HXY + BY 2 , the axes in each case being rectangular, prove that
a + b = A + B,
ab − h2 = AB − H 2
Solution: Let the axes be rotated by θ without change of origin. Let (x, y) and (X, Y ) be
coordinates of a point with referred to original and new axes. Then
x = X cos θ − Y sin θ,
y = X sin θ + Y cos θ
Substituting the values of x, y in ax2 + 2hxy + by 2 = 0
ax2 + 2hxy + by 2
= a(X cos θ − Y sin θ)2 + 2h(X cos θ − Y cos θ)(X sin θ + Y cos θ) + b(X sin θ + Y cos θ)2
= (a cos2 θ + 2h sin θ cos θ + b sin2 θ)X 2 + (a sin2 θ − 2h sin θ cos θ + b sin2 θ)Y 2 +
2[(b − a) sin θ cos θ + h(cos2 θ − sin2 θ)]XY
= AX 2 + 2HXY + BY 2
A = a cos2 θ + 2h sin θ cos θ + b sin2 θ
B = a sin2 θ − 2h sin θ cos θ + b cos2 θ
H = (b − a) sin θ cos θ + h(cos2 θ − sin2 θ)
Now
A + B = a cos2 θ + 2h sin θ cos θ + b sin2 θ + a sin2 θ − 2h sin θ cos θ + b sin2 θ
or, A + B = a(cos2 θ + sin2 θ) + b(cos2 θ + sin2 θ) = a + b
Again, lengthy calculation we will get
AB + H 2 = ab − h2
Notes: The quantities a + b and ab − h2 are called invarients of ax2 + 2hxy + by 2 .
Example 25 Prove that g 2 + f 2 is an invariant of ax2 + 2hxy + by 2 + 2gx + 2f y + c = 0 under
a rotation of axes without changing the origin.
Solution: Let θ be the angle between the old and new axes. Then
x = X cos θ − Y sin θ,
y = X sin θ + Y cos θ
Now the given equation transformed to
a(X cos θ − Y sin θ)2 + 2h(X cos θ − Y sin θ)(X sin θ + Y cos θ) + b(X sin θ + Y cos θ)2
22
+2g(X cos θ − Y sin θ) + 2f (X sin θ + Y cos θ)c = 0
Hence the first degree terms are
and 2(f cos θ − g sin θ)Y
2(g cos θ + f sin θ)X
∴ G = g cos θ + f sin θ,
F = f cos θ − g sin θ
Now,
G2 + F 2 = g 2 cos2 θ + f 2 sin2 θ + 2f g sin θ cos θ + f 2 cos2 θ + g 2 sin2 θ − 2f g sin θ cos θ
= g 2 (sin2 θ + cos2 θ) + f 2 (sin2 θ + cos2 θ) = g 2 + f 2
Example 26 If (x, y) and (X, Y ) the be coordinates of same point referred two set of rectangular axes the same origin and if ux + vy where u and v independent of x and y, becomes
V X + U Y , show that u2 + v 2 = U 2 + V 2 .
Solution: Let θ be the angle between the new and old axes. Then
x = X cos θ − Y sin θ,
y = X sin θ + Y cos θ
Now
ux + vy = u(X cos θ − Y sin θ) + v(X sin θ + Y cos θ)
= (u cos θ + v sin θ) + (v cos θ − u sin θ)Y
By question, ux + vy becomes V X + U Y , so
V = u cos θ + v sin θ,
U = v cos θ − u sin θ
Squaring and adding,
V 2 + U 2 = (u cos θ + v sin θ)2 + (v cos θ − u sin θ)2
= u2 cos2 θ + 2uv sin θ cos θ + v 2 sin2 θ + v 2 cos2 θ − 2uv sin θ cos θ + u2 sin2 θ
= u2 (sin2 θ + cos2 θ) + v 2 (sin2 θ + cos2 θ) = u2 + v 2 .
Example 27 Prove that the transformation which convert
X2 Y 2
+
into ax2 + 2hxy + by 2
p
q
will convert
X2
Y2
ax2 + 2hxy + by 2 − λ(ab − h2 )(x2 + y 2 )
+
to
p−λ q−λ
1 − λ(a + b) + λ2 (ab − h2 )
23
Solution: Let us supose the rotation of axes by θ will change
X2 Y 2
+
into ax2 + 2hxy + by 2
p
q
Then
X = x cos θ − y sin θ
Y = x sin θ + y cos θ
X 2 + Y 2 = (x cos θ − y sin θ)2 + (x sin θ + y cos θ)2
= x2 cos2 θ − 2 cos θ cos θ + y 2 sin2 θ + x2 sin2 θ + 2xy sin θ cos θ + y 2 cos2 θ
= x2 (sin2 θ + cos2 θ) + y 2 (sin2 θ + cos2 θ) = x2 + y 2
∴ X 2 + Y 2 = x2 + y 2
(38)
Again
X2 Y 2
+
=
p
q
=
=
=
(x cos θ − y sin θ)2 (x sin θ + y cos θ)2
+
p
q
2
2
2
x cos θ − 2 cos θ sin θxy + y sin2 θ
p
2
2
x sin θ + 2 cos θ sin θxy + y 2 cos2 θ
q
2
2
sin θ cos θ sin θ cos θ
cos θ sin θ
2
+
x +2
−
xy +
p
q
q
p
2
sin θ cos2 θ
+
y2
p
q
= ax2 + 2hxy + by 2
where
cos2 θ sin2 θ
+
a=
p
q
sin θ cos θ sin θ cos θ
1 1
h=
−
=−
−
sin θ cos θ
q
p
p q
sin2 θ cos2 θ
+
b=
p
q
From (39) and (41)
cos2 θ sin2 θ sin2 θ cos2 θ
a+b=
+
+
+
p
q
p
q
1
1 1
1
sin2 θ + cos2 θ + (sin2 θ + cos2 θ) = +
=
p
q
p q
24
(39)
(40)
(41)
Again,
ab =
cos2 θ sin2 θ
+
p
q
sin2 θ cos2 θ
+
p
q
sin2 θ cos2 θ cos4 θ sin4 θ sin2 θ cos2 θ
+
+
+
=
p2
pq
pq
q2
ab =
h2 =
=
=
1
1
1
2
2
sin
θ
cos
θ
+
(cos4 θ + sin4 θ)
+
2
2
p
q
pq
2
1 1
−
sin2 θ cos2 θ
p q
1
1
2
sin2 θ cos2 θ
+
−
p2 q 2 pq
1
1
2
2
2
+
sin
θ
cos
θ
−
sin2 θ cos2 θ
2
2
p
q
pq
(42)
(43)
Subtracting (43) from (42), we get
ab − h2 =
1
(cos4 θ + sin4 θ + 2 sin2 θ cos2 θ)
pq
or, ab − h2 =
1
1
(cos2 θ + sin2 θ)2 =
pq
pq
Thus,
X2 Y 2
+
convert into ax2 + 2hxy + by 2
p
q
using invariants
1 1
1
+ = a + b and
= ab − h2
p q
pq
Using these invarients
X2
Y2
+
p−λ q−λ
But we have
1 1
+ = a + b,
p q
(q − λ)X 2 + (p − λ)Y 2
(p − λ)(q − λ)
2
qX + pY 2 − λ(X 2 + Y 2 )
=
pq − λ(p + q) + λ2
2
X
Y2
λ
+
− pq
(X 2 + Y 2 )
p
q
=
2
1 − λ p1 + 1q + λpq
=
1
= ab − h2 , X 2 + Y 2 = x2 + y 2
pq
25
(44)
X2 Y 2
+
= ax2 + 2hxy + by 2
p
q
Using these results in (44)
X2
Y2
ax2 + 2hxy + by 2 − λ(ab − h2 )(X 2 + Y 2 )
+
=
p−λ q−λ
1 − λ(a + b) + (ab − h2 )λ2
26