help me out.......

this is a passage question..........i solve d the rest of them but cant do this one.....
A point is moving on curve S such that at any instant the slope is proportional to ratio of absicca with ordinate of point and area bounded by curve is finite and is equal to 4Ï€ .with respect to any ponit P on curve S,a chord of contact is drawn on x2+y2=2 which touches circle x2+y2=a

QIf curve S intersects X axis at 2 points (r1,0),(r2,0) r1<r2 then find value of \int_{\frac{r_{1}}{2}}^{\frac{r_{2}}{2}}{\frac{2u^{332}+u^{998}+4u^{1664}sinu^{691}}{1+u^{666}}}du

25 Answers

24
eureka123 ·

shayad main galti kar raha hoon[2][2].......do minute ruk....[12]

24
eureka123 ·

@ deepanshul..chord of contact not used here.....it was for other questions.......[1]

33
Abhishek Priyam ·

waise we should solve it having k it will not make much difference... only limits will differ..

33
Abhishek Priyam ·

well...k=1

option might have been free of any constant...

and that chord one is not for this q it was a complete passage...

13
deepanshu001 agarwal ·

can sum1 plz temme how u ppl hav used the second condition given about chord of contact...

and hows k=1

33
Abhishek Priyam ·

[3]

24
eureka123 ·

ooooo............mere dimag mein ye baat nahin aayi.shayad tab main coordinate ke bare mein soch raha thaa aur integration bhool gaya thaa...[3][3]

33
Abhishek Priyam ·

x ko -1+1-x se replace karke...

24
eureka123 ·

how did u simplify the integral????????????
post #13

33
Abhishek Priyam ·

in that putting 1+x666=t

sayad ye aega
I=0∫1(1+t)dt/t√t-1

11
Subash ·

okies :)

33
Abhishek Priyam ·

slope of curve means??

slope of tangent at that point...

11
Subash ·

@priyam how did you decide it was the slope of the tangent?[78]

1
Optimus Prime ·

what a complcated sum

33
Abhishek Priyam ·

So question ye hai...

I=-1∫12u332+u998/(1+u666)

I=20∫12u332+u998/(1+u666)
also 332+666=998 :P kuch hona chahiye isse.. :p
___(edited)
in that putting 1+x666=t

sayad ye aega
I=0∫1(1+t)dt/t√t-1

33
Abhishek Priyam ·

k=1 kaise aaya...

do constant hai naa..

dy/dx=kx/y ..... yaha par 1 liya kya... k=1

and one consntant from integration... and only one condition... area given so ek eliminate kaise hua...

1
MATRIX ·

hmmmmm....[55][55][55]......................................

24
eureka123 ·

@ subash edited line was ""If curve S intersects X axis at 2 points (r1,0),(r2,0) r1<r2 then find value of ""[1][1]

24
eureka123 ·

@ priyam u r rite.almost....
eqn of S aa rahi hai x2+y2=4

24
eureka123 ·

arre bhai.......yehi to hai paragraph....""A point is moving on curve S such that at any instant the slope is proportional to ratio of absicca with ordinate of point and area bounded by curve is finite and is equal to 4Ï€ .with respect to any ponit P on curve S,a chord of contact is drawn on x2+y2=2 which touches circle x2+y2=a""

33
Abhishek Priyam ·

Eureka eqn of S to nikaala hoga naa..

yahi aaya

y2+kx2=c

k>0

c=4√k

y2+kx2=4√k

11
Subash ·

erueka its a bit confusing to look at

please post the entire paragraph!

24
eureka123 ·

i missed a line in question....edited that

@ manipal ..this question has nothing to do with other questions of paragraph.........

11
Mani Pal Singh ·

if this is a paragraph question
then please post the whole paragraph
so that whole problem may be looked at !!!!!!!!!!!!!

24
eureka123 ·

thats why i cant get it............other parts of this paragraph were also very good............if any wants them then feel free to ask.......but help me here........

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