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If P(3,2) is one end of the focal chord PQ of the parabola y^2+ 4x + 4y =0, then the slope of the normal at Q is:
(A) 1/2. (B) 2. (C) 1/2. (D) 2



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shubhamsrg
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Whats your progress ?

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\[\LARGE (y+2)^{2}=4(1x) \]
so vertex=(1,2)

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dw:1359794353016:dw

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this diagram is not possible :/ acc to what im thinking..

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how can focus be (1,0)

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\[\LARGE 2y \frac{dy}{dx}+4+4\frac{dy}{dx}=0\]
i do everything i know :P because i dont know exactly what to do :o

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\[\LARGE \frac{dy}{dx}=\frac{4}{2y+4}\]

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that is for tangent..

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for normal it will be ..
\[\LARGE \frac{2y+4}{4}=\frac{8}{4}\] =2?

shubhamsrg
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okay, vertex is (1,2)

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lol

shubhamsrg
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focus would have been (1,0) if vertex had been (0,0)

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oh...yeah

shubhamsrg
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because of this shifting, new vertex = ?

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so (1+a,2)

shubhamsrg
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i mean new focus= ?

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!!!

shubhamsrg
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shift (1,0) 1 unit to the right and 2 units down

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2,2 right?

shubhamsrg
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yep,

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that's what i wrote earlier :P

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(1+a,2)
a=1

shubhamsrg
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now you have eqn of line PQ

shubhamsrg
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find coordinate of Q

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m=4

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slope of line PQ=4

shubhamsrg
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noooo.....

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:o

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\[y2=m(x+3)\]

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22=m(1)
m=4

shubhamsrg
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P(3,2)
F(2,2)
slope = 4/5

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:

shubhamsrg
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x=2 and not 2

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oh :p

shubhamsrg
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so slope = 4/5
eqn of line = ?

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5y+4x+2=0

shubhamsrg
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okay.

shubhamsrg
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eqn of parabola = y^2 + 4x+ 4y =0
eqn of line = 5y + 4x + 2 =0
solve both to solve for Q

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oh pellet :

shubhamsrg
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Note that you'll obtain 2 solutions from here, 1 should easily be (3,2)

shubhamsrg
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So its easy, form a quadratic, use sum of roots thing, you already know (3,2) is a solution, just find other coordinate

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abe yaar short method bata :P i took the wrong road nvm

shubhamsrg
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Arey ho to gaya hai, bas Q nikalna hai,

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why are we solving both the equations together? Q doesn't lie on parabola does it?

shubhamsrg
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Ofcorse it does! PQ is the focal chord

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but P doesn't lie on parabola?

shubhamsrg
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Of course it does, what are you saying!

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check it

shubhamsrg
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Just did,

shubhamsrg
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(y+2)^2 = 4(1x)

shubhamsrg
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x=3 and y=2

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but logically when I saw..
Vertex=(1,2)
P=(3,2)
how big is the parabola :X

shubhamsrg
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vertex = (1,2)

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wahi wahi

shubhamsrg
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are you sleep deprived? o.O

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:/

shubhamsrg
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I don;t understand your query

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dw:1359795719275:dw
how does it lie on the parabola?
forget the substi. method for a while

shubhamsrg
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dw:1359795779003:dw

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:O

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that would happen if
\[\LARGE y^2=4ax\]

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oh wahi to hai NVM

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sorry :/

shubhamsrg
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hmm.

shubhamsrg
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I minded! :

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sleep deprived :P

shubhamsrg
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So, shall we find Q now ?

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finding

shubhamsrg
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just forma quad, rest is easy

shubhamsrg
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y^2 + 4x+ 4y =0
5y + 4x + 2 =0
=>4x = 5y 2
make that substitution

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\[\LARGE y^2y2=0\]

shubhamsrg
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Okay, one root of this is 2,right ?

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yeah!

shubhamsrg
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use sum of roots or any other legit method, find other root.

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1

shubhamsrg
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okay, so x=?

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1?

shubhamsrg
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x= 5y2

shubhamsrg
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y=1

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oh yeah wahi :P

shubhamsrg
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i mean 4x= 5y2

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3/4

shubhamsrg
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Q(3,4 ,1)

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3/4*

shubhamsrg
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actually we needn;t find x,

shubhamsrg
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slope of normal = (2y+4)/4

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how?

shubhamsrg
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tumhi ne to upar differentiate karke nikala tha

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i dont want to live on this planet anymore

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why did we do all this then?

shubhamsrg
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We had to find y coordinate of Q

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oo I SEE

shubhamsrg
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So ans ?

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wait :o

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4 hai ya +4?

shubhamsrg
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tum batao.

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\[\LARGE \frac{2y+4}{4}\]

shubhamsrg
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4/(2y+4)
ye tangent ka hai, right?

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yeah

shubhamsrg
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slope of tangent * slope of normal = 1

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oh..okay

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so 1/2?

shubhamsrg
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abhi bhi confidence nahi aya?

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1/2.

shubhamsrg
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B

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dont say 1/2 dil tut jaega :'(

shubhamsrg
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okay, nahi kahunga

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answer to wahi hai be :P
1/2

shubhamsrg
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na, ans 1/2 hoga.

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1/2
IIT q tha ye

shubhamsrg
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leme check.


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Solution(A)

shubhamsrg
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usme kya uqes hai aur tumne kya copy kiya hai! :3

shubhamsrg
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ques*

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O_o usme ques kya hai ?

shubhamsrg
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parabola ka eqn dekho

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1/2 in my register

shubhamsrg
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abe parabola ka eqn dekho

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mine is correct :

shubhamsrg
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y^2 + 4x + 4y =0 tumne likha hai
y^2 + 4x + 4 =0 hai

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\[\LARGE \frac{4}{4+4}\]
that s what i got in last in register

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myequation is right!!
101% ques might be different..

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issi Q ke hisb se 1/2 aya hai

shubhamsrg
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ha, uska ans 1/2 hoga, lekin iska 1/2 hai.

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iska 1/2 hai

shubhamsrg
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kon se saal ka hai ?

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no idea,disha me nahi hai

shubhamsrg
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1/2 hai iska.

shubhamsrg
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1/2 uska hoga.

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IDK..its 1/2..chod

shubhamsrg
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mujhe to nahi lag raha abhi tak kahi koi galti ki hai ?

shubhamsrg
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ohh. ::O

shubhamsrg
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focus !

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o.o

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focus?

shubhamsrg
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(y+2)^2 = 4(1)((x1)
vertex = (1,2)
focus was (1,0) for vertex (0,0)
so new focus will be
(0,2)

shubhamsrg
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eqn of line =
(y2) = m(x+3)

shubhamsrg
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m= 4/3


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wait i didnt get the focus thing :/

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kya kiya tune usme?

shubhamsrg
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focus (1,0) hoga y^2 = 4ax me

shubhamsrg
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ham (1,0) like kar rahe the

shubhamsrg
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3y + 4x +6 =0

shubhamsrg
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y^2 + 4x + 4y =0
and
4x = 63y
=> y^2 + y 6 =0
1 ans hai 2,
dusra ho jayega 3

shubhamsrg
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ab slope of normal = 4/(2y+4) = 2
hahahha

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lol o.o

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par kya hai ke na focus wala part smaj nahi aya :P

shubhamsrg
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ye ques repost karo firse, koi expert batayega, ise close kar do

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do we have a greater expert than you ?:P

shubhamsrg
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wo to dikh gaya, :3

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BOLNA dhang se focus wala bus :'(

shubhamsrg
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y^2 = 4ax ke type ka hai hamara eqn

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yo

shubhamsrg
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waha focus (1,) hota hai jab vertex (0,0) hota hai.

shubhamsrg
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yaha vertex (1,2) hai

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OH :o

shubhamsrg
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to focus (1+1 , 02) = (0,2)

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:D

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m=0 aa raha hai ?

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damwait

shubhamsrg
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wait wait, 1/2 aa to raha hai!

shubhamsrg
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yeah !

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m=4/3

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wtf :/

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3y+4x+6=0!!!

shubhamsrg
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(2y+4)/4
y=3
slope = 1/2

shubhamsrg
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last me thora pagla gaya tha mai, ho to gaya, 1/2

shubhamsrg
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1/2*

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"nahi hamara answer sahi hai :p "

shubhamsrg
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focus toot gaya tha! :

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mai kar raha hoon dubara bus hogaya..1m..BC 1 ghanta hogya :/

shubhamsrg
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hmm..

shubhamsrg
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ja raha hu, bye

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y=3 hai?

shubhamsrg
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3

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yeah wahi :P

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OMG AAGAYA O____O

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ahaaa maza agaya :P mast Q :P

shubhamsrg
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aasan tha! :

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yeah :o

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chal lunch:P
thanda hogaya :/

shubhamsrg
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bye