Math 4121
Three subjects in this course – Geometry, Differential Calculus and
Integral Calculus
Books:
A Text book of coordinate geometry of two and three dimensions
with vector analysis by Rahman and Bhattacharjee.
Differential Calculus by Das and Mukherjee
Integral Calculus by Das and Mukherjee
Calculus by Anton, Bivens, Davis
Geometry
Two dimensional co-ordinate geometry: Change of axes.
Transformation of co-ordinates. Simplifications of equation of
curves.
Change of Axes
Consider a set of axes in a plane
y
P(x,y)
1
Q(2,1)
o
2
Fig1
Consider now two sets of axes in a plane
x
X
Y
y
1
o
(2,1)1
(0,0)
O
2
Fig2
(3,2)
(1,1)
Q
1
1
2
1
2
P(x,y)
2 (X,Y)
1
2
X
x
A point has one set of coordinates (x,y) for one set of axes
A point has two sets of coordinates (x,y)and (X,Y) for two sets of
axes
A point has three sets of coordinates (x,y), (X,Y) and ( x ¢ , y ¢ ), for
three sets of axes and so on.
Again suppose that we have a set of axes in a plane
y
(x-2)2+(y-1)2=1
C(2,1)
1
x
o
2
Fig3
2
Suppose that we have two sets of axes in a plane
Y
(x-3)2+(y-2)2=1
(X-1)2+(Y-1)2=1
y
(2,1) 1
(0,0) O
1
o
(3,2)
(1,1)
C
1
12
P(x,y)
(X,Y)
2
1 2+(y-1)2=1
(x-2)
22+Y2=1
X
X
x
Fig4
A curve has one equation for one set of axes
A curve has two equations for two sets of axes
A curve has three equations for three sets of axes and so on
We see that there is a relation between two systems of coordinates
i.e. (x,y) and (X,Y) system x=X+2, y=Y+1
We shall now derive a general formula which will give the
relation between (x,y) and (X,Y)
Transformation of coordinates
The process of changing the coordinate of a point or the equation
of a curve is called transformation of coordinates. This
transformation is performed through change of axes. There are
three types of transformations.
1.Translation of axes (origin changed but the direction of axes
unchanged)
2.Rotation of axes (origin not changed but the direction of axes
changed)
3
x
3.Translation and rotation of axes (both the origin and the direction
of axes changed)
1. Translation of axes
Y
y
P(x,y)
(X,Y)
(a , b )
o
(0,0) O
Oo
a
C
X
X
Y
y
A
X
b
x
B1
2
Fig5
x=OB=OC+CB=X+ a
y=PB=PA+AB=Y+ b
If we transfer the origin at the point ( a , b ) the equations of
transformation are
x=X+ a , y=Y+ b
2. Rotation of axes
(x,y)
P (X,Y)
Y
XY
xq
y
M1
O
X
x
q
x
Fig62
N1
y
M N
x x
2
4
X
x
2
x
x=OM=ON-MN=ON- M ¢N ¢ =Xcos q -Ysin q
y=PM= MM ¢ + PM ¢ = NN ¢ + PM ¢ = Xsin q +Ycos q
If we rotate the axes through an angle q the equations of
transformations are
x=Xcos q -Ysin q
y= Xsin q +Ycos q
3. Translation and rotation of axes
y1
P(x,y)
(X,Y)
Y
(x1,y1)
y
(a , b )
o
(0,0) O
Oo
q
x1
X
x
x
Fig7
x=X+ a = a + x ¢ cos q - y ¢ sin q
y=Y+ b = b + x ¢ sin q + y ¢ cos q
Simplification of equation of curves by transformation of
coordinates
The general equation of second degree in x and y
is ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
We can simplify the equation by transformation of the coordinates.
x and y term can be removed by translation of axes, xy term can be
removed by rotation of axes.
Removal of x and y term from ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
5
Let us transfer the origin to the point ( a , b ) with a new set of axes
then the equations of transformation are x=X+ a , y=Y+ b
The equation becomes
a(X+ a )2+2h (X+ a )(Y+ b )+b(Y+ b )2+2g (X+ a )+2f(Y+ b )+c=0
aX2+2hXY+bY2+2(a a +h b +g)X+2(h a +b b +f)Y+
aa 2 + 2hab + bb 2 + 2 ga + 2 fb + c = 0
To remove x and y term put
a a +h b +g=0
h a +b b +f=0
a
hf - bg
a=
=
b
hg - af
hf - bg
,
ab - h 2
=
1
ab - h 2
b=
hg - af
ab - h 2
Now the equation becomes
aX2+2hXY+bY2+ aa 2 + 2hab + bb 2 + 2 ga + 2 fb + c = 0
aX2+2hXY+bY2+ a (a a +h b +g)+ b (h a +b b +f)+g a +f b +c=0
aX2+2hXY+bY2+g a +f b +c=0
If
we
transfer
the
origin
at
(a , b )
the
equation
ax + 2hxy + by + 2 gx + 2 fy + c = 0
2
2
becomes aX2+2hXY+bY2+g a +f b +c=0
Example: Transform the equation 3x 2 + 2 xy + 3 y 2 - 18 x - 22 y + 50 = 0 to
one in which there is no term involving x and y
a=3, b=3, c=50, f=-11, g=-9, h=1
a=
hf - bg
= 2,
ab - h 2
b=
hg - af
=3
ab - h 2
Now let us transfer the origin at (2,3) then the equations of
transformations are
x=X+2, y=Y+3
6
therefore the equation becomes
3(X+2)2+2 (X+2)(Y+3)+3(Y+3)2-18 (X+2)-22(Y+3)+50=0
………………………………….
3X2+2XY+3Y2 -1=0
3X2+2XY+3Y2 =1
therefore the equation becomes
3X2+2XY+3Y2 + g a +f b +c=0
3X2+2XY+3Y2 +(-9) 2+(-11)3+50=0
3X2+2XY+3Y2 -1=0
3X2+2XY+3Y2 =1
07-06-21CA
08-06-21B
Invariants of transformation
Suppose the equation is ax 2 + 2hxy + by 2 = c
If we rotate the axes through an angle q the equations of
transformations are
x=Xcos q -Ysin q
y= Xsin q +Ycos q
Now
a (Xcos q -Ysin q )2+2h (Xcos q -Ysin q )( Xsin q +Ycos q )
+b (Xsin q +Ycos q )2 = c
(acos2 q +2hsin q cos q +bsin2 q )X2 +2(h(cos2 q -sin2 q )-(a-b)
sin q cos q ))XY+(asin2 q -2hsin q cos q +bcos2 q )Y2 =c
a ¢ X2+2 h ¢ XY+ b ¢ Y2=c
where
a ¢ = acos2 q +2hsin q cos q +bsin2 q
b ¢ = asin2 q -2hsin q cos q +bcos2 q
h ¢ = h(cos2 q -sin2 q )-(a-b) sin q cos q
a ¢ + b ¢ =a+b
7
2 a ¢ = 2acos2 q +4hsin q cos q +2bsin2 q
=a(1+cos2 q ) +2hsin2 q +b(1-cos2 q )
=(a+b)+((2hsin2 q +(a-b)cos2 q )
2 b¢ = (a+b)-((2hsin2 q +(a-b)cos2 q )
2 h¢ = 2hcos2 q -(a-b) sin2 q
4a ¢b ¢ - 4h ¢ 2
=(a+b)2-((2hsin2 q +(a-b)cos2 q )2-(2hcos2 q -(a-b)
sin2 q )2
=(a+b)2-4h 2 -(a-b)2
=4ab-4h 2
a ¢b ¢ - h ¢ 2 =ab-h 2
Invariants of transformation are
a ¢ + b ¢ =a+b,
a ¢b ¢ - h ¢ 2 =ab-h 2
Removal of xy term
h¢ = 0
h(cos2 q -sin2 q )-(a-b) sin q cos q =0
2hcos2 q -(a-b) sin2 q =0
tan2 q = 2h or, q = 1 tan -1 2h
a-b
2
a-b
Ex. Remove the xy term from the equation 7x2-6 3 xy+13y2=16
and find the equations of transformations
2 tan q
=-6 3
2
a-b
1 - tan q 7 - 13
tan q = - 2 ± 4 + 12 or, 1 , - 3
2 3
3
tan q = 1 , q = p
6
3
tan2 q = 2h
or,
or,
tan q = - 3 = - tan p = tan ( p - p )
3
3 tan2 q +2tan q - 3 =0
or, q = 2p
3
3
8
1
2
q = tan -1
2h
= 1 tan -1 - 6 3 = 1 tan -1 3 = 1 p = p , we get one angle
2 3 6
7 - 13 2
a-b 2
For q = p
6
x=Xcos q -Ysin q =X 3 -Y 1 = 1 ( 3 X-Y)
2
2 2
1
y= Xsin q +Ycos q = (X+ 3 Y)
2
For, q = 2p
3
x=Xcos q -Ysin q = - 1 (X+ 3 Y)
2
1
y= Xsin q +Ycos q = ( 3 X-Y)
2
For q = p
2
7x -6
6
3 xy+13y2=16
7
( 3 X-Y)2- 6 3 ( 3 X-Y) (X+ 3 Y)+ 13 (X+ 3 Y)2=16
4
4
4
……………………
………………………
X2+4Y2=4
Y2
X2
or, 2 + 2 = 1
2
1
ellipse
Similarly For, q = 2p
3
2
2
4X +Y =4
2
2
or, X2 + Y 2 = 1
1
2
ellipse
The equations of transformations are
x= 1 ( 3 X-Y)
2
9
y= 1 (X+ 3 Y)
2
and
x= - 1 (X+ 3 Y)
2
1
y= ( 3 X-Y)
2
Y
y
X
X
2
q = 120
2 q = 30
1
o1
x
ellipse
Y
Fig8
y
o
10
x
ellipse
y
x
o
ellipse
x2 y2
+
=1
a2 b2
Alternative method
2 tan q
= -6 3 = 3
2
a-b
1 - tan q 7 - 13
tan q = - 2 ± 4 + 12 or, 1 , - 3
2 3
3
tan q = 1 , q = p
6
3
tan2 q = 2h
or,
tan q = - 3 = - tan p = tan ( p - p )
3
3
or,
or, q = 2p
3
For q = p
6
x=Xcos q -Ysin q =X 3 -Y 1 = 1 ( 3 X-Y)
2
2
2
y= Xsin q +Ycos q = 1 (X+ 3 Y)
For, q = 2p
3
3 tan2 q +2tan q - 3 =0
2
11
x=Xcos q -Ysin q = - 1 (X+ 3 Y)
2
1
y= Xsin q +Ycos q = ( 3 X-Y)
2
The equation is
a ¢ X2+ b ¢ Y2=16 where
a ¢ + b ¢ =a+b =7+13 =20,
a ¢b ¢ =ab-h 2 =91-27=64
a ¢ - b ¢ = ± 12
( a ¢ - b ¢ )2=( a ¢ + b ¢ )2-4 a ¢b ¢ =144
taking + sign a ¢ = 16 b ¢ = 4
b ¢ = 16
taking - sign a ¢ = 4
16X2+4Y2=16
or, 4X2+Y2=4
2
2
or, X2 + Y 2 = 1
1
2
ellipse
4X2+16Y2=16
or, X2+4Y2=4
2
2
or, X 2 + Y2 = 1
2
1
ellipse
The equations of transformations are
x= 1 ( 3 X-Y)
2
y= 1 (X+ 3 Y)
2
and
x= - 1 (X+ 3 Y)
2
1
y= ( 3 X-Y)
2
08-06-21A
12
Ex. Remove x, y and xy term from the following equation
17 x 2 + 18 xy - 7 y 2 - 16 x - 32 y - 18 = 0 . Also find the equations of
transformations.
a=17, b=-7, c=-18, f=-16, g=-8, h=9
a=
hf - bg
= 1,
ab - h 2
b=
hg - af
=-1
ab - h 2
Now let us transfer the origin at (1,-1) then the equations of
transformations are
x=X+1, y=Y-1
therefore the equation becomes
17X2+18XY-7Y2 + g a +f b +c=0
17X2+18XY-7Y2 + (-8)1+(-16)(-1)-18=0
17X2+18XY-7Y2 =10 …… (1)
tan2 q = 2h or, 2 tan q2 = 3 or, 3tan2 q +8tan q -3=0
a-b
1 - tan q 4
tan q = - 8 ± 64 + 36 or, 1 , - 3
6
3
tan q = 1
sin q = 1 ,
cos q = 3 ,
3
10
10
X= x ¢ cos q - y ¢ sin q = 1 ( 3 x ¢ - y¢ )
Y= x ¢ sin q + y ¢ cos q =
tan q =-3
10
1
10
( x ¢ + 3 y¢ )
sin q = 3 , cos q = - 1 ,
10
10
,
X= x ¢ cos q - y ¢ sin q = - 1 ( x ¢ +3 y¢ )
Y= x ¢ sin q + y¢ cos q =
10
1
10
(3 x ¢ - y¢ )
The equation (1) after rotation of axes becomes
where
13
a ¢ x ¢ 2+ b ¢ y¢ 2=10
a ¢ + b ¢ =a+b =10,
a ¢b ¢ =ab-h 2 = -200
a ¢ - b ¢ = ± 30
( a ¢ - b ¢ )2=( a ¢ + b ¢ )2-4 a ¢b ¢ =900
b ¢ = -10
taking + sign a ¢ = 20
b ¢ = 20
taking - sign a ¢ = -10
20 x ¢ 2-10 y¢ 2=10
or, 2 x ¢ 2- y¢ 2= 1
or,
x¢ 2
y¢2
=1
12
2
æ 1 ö
ç
÷
è 2ø
-10 x ¢ 2+20 y¢ 2=10
hyperbola
or, x ¢ 2-2 y¢ 2= -1
y¢2
x¢2
or, - 2 +
= 1 hyperbola
2
1
æ 1 ö
ç
÷
è 2ø
The equations of transformations are
x=X+1= 1 ( 3 x ¢ - y¢ )+1
10
y=Y-1=
1
10
( x ¢ + 3 y¢ )-1
and
x=X+1= - 1 ( x ¢ +3 y¢ )+1
10
y=Y-1=
1
10
(3 x ¢ - y¢ )-1
y
y1
Y
14
x1
x1
q = 108.43
q = 18.43
o
O(1,-1)
y1
x
X
hyperbola
Fig9
09-06-21BC
Pair of straight lines
Suppose x+y=0 and x-y=0 are two straight lines
Then (x+y)(x-y)=0 or, x2-y2=0 is a pair of straight lines
Again x+2y+3=0 and 2x+3y +1=0 are two straight lines
Then (x+2y+3)(2x+3y+1)=0
or, 2x2+7xy+6y2+7x+11y+3=0 is a pair of straight lines
Theorem: Find the condition for which the general equation of
second degree ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 represents a pair of
straight lines.
ax2+2(hy+g)x+by2+2fy+c=0 ……..(1)
x=
- 2(hy + g ) ± 4(hy + g ) 2 - 4a (by 2 + 2 fy + c)
=
2a
- (hy + g ) ± (hy + g ) 2 - a (by 2 + 2 fy + c)
=
a
- (hy + g ) ± (h - ab) y 2 - 2(hg - af ) y + g 2 - ac)
2
a
For the equation (1) to represent a pair of straight lines
15
the quantity under the square root must be a perfect square,
therefore
4(hg-af)2 = 4(h2-ab)(g2-ac)
abc+2fgh-af2-bg2-ch2=0
a(bc-f2)-h(hc-fg)+g(hf-bg)=0
a
h
g
h
b
f =0.
g
f
c
D =0
a
h
g
where D = h
b
f
g
f
c
Ex. Show that the equation 6 x 2 - 5 xy - 6 y 2 + 14 x + 5 y + 4 = 0 represents a
pair of straight lines. Find the separate equations of the straight
lines.
a=6, b=-6, c=4, f=5/2, g=7, h=-5/2
-5/ 2
7
f = -5/ 2
-6
5 / 2 =0
c
5/ 2
4
a
h
g
First show that D = h
b
g
f
6
7
Now 6x2-(5y-14)x-6y2+5y+4=0
x=
5 y - 14 ± (5 y - 14) 2 - 4.6(-6 y 2 + 5 y + 4)
12
= 5 y - 14 ± (169 y - 260 y + 100)
2
= 18 y - 24 ,
12
12
- 8y - 4
12
12x-18y+24=0
or, 2x-3y+4=0
12x+8y+4=0
or, 3x+2y+1=0
16
x
y
1
2
4
4
7
6
x
y
1
-2
3
-5
5
-8
Ex.
of
k for which the equation
kx + 4 xy + y - 4 x - 2 y - 3 = 0 represents a pair of straight lines
2
Find
the
value
2
Different conditions for which
represents different conics/curves
a
h
g
Suppose D = h
b
f
g
f
c
ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
The equation ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0 represents
1. a pair of straight lines if D =0
2. a circle if D ¹ 0, a=b and h=0
3. a parabola if D ¹ 0 and h2=ab
4. an ellipse if D ¹ 0 and h2 < ab
5. a hyperbola if D ¹ 0 and h2 > ab
Parabola
17
Ex. Identify the curve 16 x 2 - 24 xy + 9 y 2 - 104 x - 172 y + 44 = 0 . Reduce it
to standard form and give a rough sketch of the curve.
16.9=(-12)2
a parabola
D ¹ 0 and h2=ab
2
2
16 x - 24 xy + 9 y - 104 x - 172 y + 44 = 0 .
(4x-3y)2= 104x+172y-44
The lines 4x-3y=0 and 104x+172y-44=0 are not at right angles
hence let us introduce a constant k
(4x-3y+k)2= 104x+172y-44+2k(4x-3y)+k2
= (104+8k)x+(172-6k)y+ k2-44
The lines 4x-3y+k=0 and
(104+8k)x+(172-6k)y+ k2-44=0
are at right angles if a1a2+b1b2=0
or, 4(104+8k)+(-3)(172-6k)=0
k=2
Explanation
y
x
x=0
(x,y)
y
o
y=0
x
y 2 = 4ax
y=0, x-axis
x=0, y-axis
They are perpendicular to each other.
In the coordinates (x,y),
x is perpendicular distance from y-axis
and y is perpendicular distance from x-axis
18
Now
(4x-3y+2)2= 120x+160y-40
(4x-3y+2)2= 40(3x+4y-1)
2
æ
ö
æ 4x - 3y + 2 ö
ç
÷ 25 = 40.5 ç 3 x + 4 y - 1 ÷
ç
ç
2
2 ÷
2
2 ÷
è 3 +4 ø
è 4 +3 ø
2
æ
ö
æ 4x - 3y + 2 ö
ç
÷ = 8 ç 3x + 4 y - 1 ÷
ç
ç
2
2 ÷
2
2 ÷
è 3 +4 ø
è 4 +3 ø
4x - 3y + 2
2
Y =8X where Y =
42 + 3
X= 3x +2 4 y -2 1
,
2
3 +4
2
Y =4AX
14-06-21 A
X axis Y=0, 4 x - 3 y + 2 =0
Y axis X=0, 3x + 4 y - 1=0
Vertex (0,0) X=0, Y=0
3 x + 4 y - 1=0
4 x - 3 y + 2 =0
æ 1 2ö
ç- , ÷
è 5 5ø
Focus (A,0) X=A, Y=0
3x + 4 y - 1
=2 or, 3x+4y-1=0
2
2
3 +4
4x - 3y + 2
4 2 + 32
=0 or, 4x-3y+2=0 (1,2)
A=2 (distance between vertex and focus)
semi latus rectum=2A=4
Parabola cuts on the x-axis where 16 x 2 - 104 x + 44 = 0 , x=0.455, 6.045.
Parabola cuts on the y-axis where 9 y 2 - 172 y + 44 = 0 , y=0.26, 18.85.
(0, 18.85)
19
y
Y
X
x
4
X-axis, Y=0
4x-3y+2=0
3
2
2
Y-axis, X=0
3x+4y-1=0
(1,2)
2
1
(-1/5, 2/5)
-1
2
O
xo
(0, 0.26)
(0.455,0)
2/3)
1
2
3
2
2
2
(6.045, 0)
4
5
2
6
2
Fig10
End
14-6-21 C
Problems to be done
1. Remove x, y and xy term from the following equations. Also
find the equations of transformations.
2. Identify the curve and reduce it to standard form.
3. Show that the following equations represent a pair of straight
lines. Find the separate equation of straight lines.
1. 7x2-6 3 xy+13y2=16
ellipse
2. 17 x 2 + 18 xy - 7 y 2 - 16 x - 32 y - 18 = 0 hyperbola
3. 4 x 2 + 4 y 2 + 16 x - 16 y + 7 = 0 circle
20
x
4. 3x 2 + 2 xy + 3 y 2 - 18 x - 22 y + 50 = 0 hyperbola
5. 5 x 2 + 4 xy + 2 y 2 - 2 x + 4 y + 4 = 0
6. xy - 2 x + y - 6 = 0
7. 9 x 2 + 16 y 2 - 108 x + 128 y + 256 = 0
8. 36 x 2 + 24 xy + 29 y 2 - 72 x + 126 y + 81 = 0
9. 5 x 2 - 24 xy - 5 y 2 + 4 x + 58 y + 59 = 0
10. 19 x 2 + 5 xy + 7 y 2 = 13
11. 3x 2 + 2 xy + 3 y 2 = 2
12. 6 x 2 - 5 xy - 6 y 2 + 14 x + 5 y + 4 = 0 pair of straight lines
13. 3x 2 - 8 xy - 3 y 2 - 29 x + 3 y - 18 = 0
14. 4 x 2 - 4 xy + y 2 - 8 x - y + 6 = 0 Parabola
15. x 2 + 2 xy + y 2 - 2 x - 1 = 0 parabola
16. 16 x 2 - 24 xy + 9 y 2 - 104 x - 172 y + 44 = 0 . Parabola
Section A
Math 4121
Date 29-06-21
Identify the curve 49 x 2 - 56 xy + 16 y 2 + 30 x - 45 y + 3 = 0 and reduce it
to standard form.
21