1addition : i want a short method! to solve in three mins.
if the line y-√3x-3=0 cuts the parabola y2=x+2 at A and B, and if P=(√3,0), then PA.PB is equal to …?
a) 2(3^0.5 +2)/3
b) 4(3^0.5)/2
c) 4(2 - 3^0.5)/3
d) 4(3^0.5 + 2)/3
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9 Answers
1Is it c?
the symmetric form gives
x = (r/2) + √3
y= (√3)r -2 .
substituting in the eqn of parabola , I got r1.r2 = ( 2 -√3) / 3 . so I go with c, even though there is a factor 4 missing.
Please tell where I'm wrong.
62should it not be
x=-√3+r/2
y = r√3/2
Your method is gr8.. i seem to have lost it :)
62the answer is indeed c.. but nothing is missing..
good method srinath and good question sky. :)
1the symmetric form is , well
(x -x1)/ cosθ = (y-y1)/sinθ = r
right ? the line passes through (√3 , 0 )
so x1 = √3 , y1=0
so tanθ = √3 , given . so x= (r/2) +√3
and y= r√3 /2
so solving we ,get.
r1.r2 is (2+√3) 4/3 i.e. D
62i think there is a n issue with this problem
It would have been the right solution srinath if u had the piont passing thru this line..
We both have mistakes in the solution...
do u figure out wwhat i am trying to say...
the symmetric form wont hold at all.. cos teh two segments PA and PB are not of the same slope as the line y-√3x-3
62This would have been a good question if the point P lied on this line..
I guess there is some typo.. sky pls check carefully and let us know..
1INDEED THE POINT LIES ON THE LINE!
I APOLOGISE TO TYPE A WRONG QUESTION :(
THE LINE IS y-√3x+3=0.
THANK YOU FOR YOUR REPLIES! :)