ORTHOGNAL

TRAJECTORIES
ORTHOGONAL
1.
COORDINATES TRAJECTORIES IN CARTESIAN

Definition. Trajectory. A cUrve_ which cuts every

family of curves in accordance with some member of a given
trajectory of the given family of curves. given condition is called
Definition. Orthogonal Trajectory. A curve which cuts
of a given family of curves at right every member
angles is called an orthogonal
trajectory.
Working method to find Orthogonal Trajectories :
Consider a family of curves
fa,y, c) =0, ...(1)
where c is an arbitrary constant.
Theorthogonal trajectories of a given family of curves can be
as follows : found
Step I. Obtain the differential equation of given family of
by curves (1)
differentiatingequation (1) w.r.t. x and then eliminating the arbitrary
Constant c between (1) and the derived equation.
Let the resulting differential equation of family (1) be
the form expressible in
dy -F(*, y). ...(2)
Step II. The orthogonal trajectory of the given family of curves
intersects each curve of the family at right angles, therefore slope of any
orthogonal trajectory at point (*, y) is - F(*, y)
 104 PRECIZE DIFFERENTIAL
The differential equation of orthogonal trajectories can be
written as
dy
F(*,)) ..43)
The equation (3)may be obtained from (2) after replacing dy
d

Step II. Solution of (3) gives the required equation of orthogonal

trajectories of (1).
Definition. Self Orthogonal Family. If each member of a given family
of curves intersects all other members orthogonally, then the given family
of curves is said to be self orthogonal.
Remark. If the differential equation of a family of curves is identical
with the differential equation of its orthogonal trajectories, then that
family is self orthogonal.

cUSTRATEVE EXAMPLES
Etample 1. Find the orthogonal trajectories of a family of parabolas
whose equation is y =4 ax, abeing a
parameter. (P.U. 2012, 2013)
Sol. The given equation of a family of
parabolas is
y =4 q x. .(1)
Differentiating w.r.t. x, we have
2y2de =4 a
..2)
2d
Using (2) in (1), we have
y =2xy
d
 ORTHOGONAL TRAJECTORIES 105

dr 2x [: y 0]
dyd
hich is the differential equation of given family of parabolas.
The diferential equation representing the orthogonal trajectories is Ohfad hy
2*
dx
y dy =-2x d
cy
Integrating, 2
=-+,
2
where cis an arbitrary constant.
y'+2*? =c,
which represents the required orthogonal trajectory.
Brmple 2. Find the orthogonal trajectories of the family of curves
ty=, where c is a parameter. (P.U. 2014)

Sol The given family of curves is xyc, ..(1)

Differentiating w.r.t. x, weget
2xy+ 29 -0 >2 y+* =0, ..2)

which is the differential equator of given family of curves.

Replacing by , the differential equation of required
de dy
orthogonal trajectories is
=0

ie. 2y dy - x de =0.
Integrating. we get
) -, where c is an arbitrary constant.
2 2,
2y-c,
which is sthethe required equation of the orthogonalltraiectories.
 co-axi
Exanpe 3. Find orthogonal trajectories of the family of
+y²+2gx+c=0, g being aparameter. circle
(P.U. 2005, 2015;
SoL. Given family of circles is x+y+2gx +c= o.
G.N.D.U.2
Differentiating w.r.t. x, we get
2x +2yy'+2g =0
g--(*+yy').
Using (2) in (1), we get
+y'-2?- 2y'yx +t e=0
ie y² --2xy9 +e=0.

Replacing by , We get
dy

y -*+2xy +c=0
dy
2x d x² -y'-c
dy y y
Putting =, 2 ; = d, we get
dy dy
dz
dy
(y +)
which is a linear
diferential equation.
Integrating factor e ,-logy
=e

Solution of (4) is given by z-- y +e

 ORTHOG

4
=-y+ +c,
2=- y* +ct cy
2+y =c+ cy, where c, is an arbitrary constant,
which is the required equation of orthogonal trajectories.
Erample 4. Find the orthogonal trajectories of the curves y = ce-3x
where c is a paramneter. (P.U. 2013)
Sol. Given equation of family of curves is
y ce-3*
..(1)
dy =-3ce-3x
Differentiating w.r.t. x, we get,
d

dy
dx
--3y, [Using (1 )] ..2)
which is the differential equation of given family of curves.
dx
Replacing by in (2), the differential equation of orthogonal
de dy
trajectories is
d
dy =-3y
d =3 ydy i.e. 3y dy =d.
Integrating, the required equation of orthogonal trajectories is
=Itc where c isan arbitrary constant.
mple 5. Find the orthogonal trajectories of the family of curves
+ b'+ =1, where is a parameter. (G.N.D.U. 2008)
or

Show that the family of curves .+ = 1, Abeing a

ameter, is self-orthogonal. a + b+
(P.U. 2007)
 EQUATIONS-| (SEMESTER-I) (
108 PRECIZE
DIFFERENTIAL (PU)
curves is
Sol. Given family of
y =1.
a' +4 b² +
Differentiating (1)w.r.t. x, we get
2x 2y_=0
a'+1 b +1
x(6°+2)+yy' (a² +1) =0
A
(etyy)-- (b'x+ayy)

x+yy'

Now, a?+i = a? x+yy'

a'x-b'x (a-b)xes2)
x+yy' *+yy'

and *+yy'o
-(6²-a)yy (a'-b')yy' ..3)
x+yy' x+yy'
Puting values of a+1and b+1 from (2) and (3) in (1), we get
=1
(a? -b)x (a-b°)y'
X+yy'. *+yy'
x(*+yy) +yy)
a'-b? (a²-b²)y'
..(49)

which is the differential equation of given family of curves.

 ORTHOGONAL TRAJECTORIES
d
109

Replacing y' by-in (9. the differential equation of family of

orthogonal trajectories is

- (a' -b)

-*yy)- (a -s), which is same as (4).

Hence, the given family of curves is self orthogonal.
Example 6. Prove that the system of confocal and co-axial parabolas
j'=4a(rta) is self-orthogonal.
Sol. The given family of parabolas is
y =4 a *+ a). ..(1)
Differentiating (1) w.r.t. x, we have
=4a
de

dy =2a. ..2)
de
Eliminating a between (1) and (2), we get

-2yxy. ..3)

Which is the differential equation of given family of curves.

Replacing by - in (3), the equation of orthogonal trajectories is
dy
 i.e.
--2y:+y?

i.e.

which is same as (3).

Hence, the given family of parabolas is self-orthogonal.
Example 7. Find orthogonal trajectories of the family of ellipses with
centres at (0,0) and vertices at (-1,0) and (1,0).
Sol. Since centre of the family of ellipses is (0, 0), vertices (-1,0) and
(1,0) are on x-axis, therefore equation of family is

=|
L(orb)
()2 (a,0)
ie.
(o9)

where b is a parameter.
o,-b)
Differentiating (1) w.r.t. x, we have
2x+ 2y dy =0

1 xd: ..2)
y dy
Using (2) in (1), we get
x d =1
ydy
dx .3)
=1,

which is the differential equation of given family of curves.

 ORTHOGONAL TRAJECTORIES 111
de
Replacing by - in (3), we get

+xy tyd-o
ty y=0.
Integrating, we get

-log |x| + 2
=c, Where c is an arbitrary constant
2

2+y'-2 log|x| =2 c,
which represents the required orthogonal trajectories.
Example 8. Find the value of kso that the parabolas y= cx +kare the
orhogonal trajectories ofthe family of ellipses x+2y-y = d, where
cand d are parameters.
Sol. Given family of ellipses is
**+2y'-y =d. P(1)
Differentiating (1) w.r.t. x, we get
2x+4,4 -0, .2)
d de

Which is the diferential equation of given family of curves.

d
Replacing by - , We get
du dy
d =0
2x+4y dy
dx
2x+(1 -4 y) dy =0
dx dy -=0
2x 1-4y
 Integrating, we get
1
log|x-log| 1-4y|=c, where c' is an arbitrary constang
2 4

’ 2log|x|- log | 1-4y|=4c.

’ log =4c'
|1-4 |
=A, where A=+e4c'
1-4y

1-4y =A b e
1
y=-t
A 4

iLe. (y=cx+ here c=

which represents the orthogonal trajectories of given family of ellipses.
By question, orthogonal trajectories of given family is y = cxth.
Comparing both the equations, we get, k=. 1
4