3thnx for the hint....i will try now and let u khow if i still face any problem while solving the problem!
11 Answers
1357abhishek the hint is simple
p2=a2+(b2-a2)sin2θ
Now try to prove...
33i am getting stuck after reaching the following step
(dp/dθ)2+p(d2p/dθ2)=(b2-a2)SinθCosθ
can somebody plz help me out after having reached this step??
49why do u go by this way? it is just a simple problem!!! [11]
on differentiating w.r.t. θ
we get 2pdpdθ=(b2-a2)2.sinθcosθ
so, dpdθ=(b2-a2)2.sinθcosθ2p
so from here we get d2pdθ2
and then addd p to it!![1][1]
1Subho , this is a pretty pretty lengthy process if you go by this way !!!!
3I am not able to solve the problem......
can somebody help me out???
plzz
1abhisek i thnk u made a typo in the question
that shuld have been p = a2cos2θ + b2sin2θ
not p2 = a2cos2θ + b2sin2θ
[7]