33
Abhishek Priyam
·2009-01-16 03:55:31
[1]
No common normal here...
therefore by the method of "all questions given now are solvable" and "in exams there is time limit"...
I think answer will be ...
shortest distance is distance between (-10,0) and foot of perpendicular from it on line which is same as distance between (10,0) and foot of perpendicular on line..
62
Lokesh Verma
·2009-01-16 04:45:25
what can i do if people are as smart as u are celestine...
U killed my fallacy ;) :P
9
Celestine preetham
·2009-01-16 04:51:36
abishek can u be more elaborate on ur method i cudnt understand that pls explain ;)
method of "all questions given now are solvable" and "in exams there is time limit"...
9
Celestine preetham
·2009-01-16 04:53:53
but then wat fallacy is there to kill , i thouht it was straightforward
62
Lokesh Verma
·2009-01-16 04:55:44
basically FIITJEE had solved this question incorrectly
and I could see why they had done that....
They found out wrong theta.. (a sec theta, b tan theta)
So they got a wrong answer...
Basically their answer was in the 4th Quadrant! which many ppl wud get!
62
Lokesh Verma
·2009-01-16 04:56:00
not this same question.. but a different data...
9
Celestine preetham
·2009-01-16 04:59:34
yes i had to consider only one theta other one will give local minima but not global minima ;)
13
deepanshu001 agarwal
·2009-03-16 10:20:37
can u plzz help me wid dis 1.... celestine
1
°ღ•๓ÑÏ…Î
·2009-03-16 10:27:49
btw nsihu bhaiya aapne upar parabola likha hai :P