TRAJECTORIES

                TRAJECTORIES
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A curve cuts every member of a given family of curves according to a given law is called a Trajectories of the given family.
The trajectory will be called ORTHOGONAL if each trajectory cuts every member of given family at right angle.

WORKING RULE FOR FINDING ORTHOGONAL TRAJECTORY
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1) A Form the differential equation
    of family of curves.
2) Write - d x/dy for dy/ dx
    or r² dα/dr for dr/dα if
    differential equation is in the
    polar form.
3) Solve the new differential
    equation to get the equation of
    orthogonal Trajectories.

NOTE ::
A family of curves is self-orthogonal if it is its own orthogonal family.

Question 1)
Find the value of k such that the family of parabola y= cx²+k is the othogonal trajectory of the family of ellipse x²+2y²-y=c.

SOLUTION
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Differentiate both the sides of
  x²+2y²-y=c w.r.t.x we get
  2x+4y dy/dx - dy/dx =0 
or 2x+(4y -1) dy/dx =0 is the differential equation of the given family of curves.
Replacing dy/dx by  - dx/dy to obtain the differential equation of the orthogonal trajectories,
we get   2x+ (1-4y) dx/dy= 0
⇒ dy/dx= (4y-1)/2x
⇒ ∫ dy/(4y-1)  = ∫ dx/2x 
⇒1/4log(4y-1)=1/2log(x)
                                     +1/2log(a),
    where a is any constant.
⇒ log(4y-1)=2log(x)+2log(a)
     or  4y-1=a²x².
or  y=(a²x²)/4   + 1/4,
   is the required orthogonal
   trajectory, which is of the form
   y= cx²+k.
   where c=a²/4, k=1/4.

Do yourself
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1) prove x²(a²+λ) + y²/(b²+λ)  =1
    are self orthogonal family of
    curves.

2) Find the orthogonal trajectories
     of the following families of
     curves.
a) x+2y=C

b) y= C e⁻²ˣ