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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Feb 2007 15:10:10 IST
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The tangent and normal at the point P(at2,2at) to the parabola y2=4ax meet the x-axis in T and G respectively,then angle at which the tangent at P to the parabola is inclined to the tangent at P to the circle through P,T,G is?
[ans:tan-1(t)] I'm getting different ans
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Feb 2007 15:17:12 IST
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ghj
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Feb 2007 18:09:57 IST
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according to your question a tangent cuts x-axis in 2 different points is at possible??
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Feb 2007 18:16:47 IST
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neeraj i suppose,
u can find the equation of the circle by the given three points and then find the tangent due to P on the circle and then find the angle b/w them by the formula of tanQ!
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will make it BIG! |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Feb 2007 21:09:06 IST
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avdesh...I tried it but am getting different ans..
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Feb 2007 21:26:28 IST
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then ur answer is correct...anyways i think this is lengthy but sure method
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will make it BIG! |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Feb 2007 22:11:41 IST
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Hi neeraj, The circle is nothing but the 1 with diameter as TG and center at S(a,0) ; Let the  PTS =  ;{ where Tan  = 1 / t } Let the angle bt tangents be  ; So, ext angle  PSG = {  +(  /2 -  )} Just expand..Tan  PSG { = 2t / (t 2-1) } ; Solve for cot  ;{u get it as 1/t } So,  = Tan -1(1/t) ....ans. 
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Sorry,i ate ur brain!!! |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Feb 2007 14:48:44 IST
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You just draw the figure.Since tangent and normal at P are perpendicular to each other,so TPG = 900. This is the angle contained in a semicircle and hence TG would be the diameter of the circle passing through P,T and G.
In equation of tangent and normal put y=0 to get T and G. Then T is (-at2,0) and G is (2a - at2,0) Hence centre of the circle is (a,0) which is the focus of parabola.
Slope of normal to the circle at P = 2at/(at2-a) = 2t/(t2-1) Hence slope of tangent to the circle at P = -1/(slope of normal) = (1-t2)/2t
Also slope of tangent at P to parabola = 1/t
Now slopes of two lines are known and you can easily find the angle between them.
Best Wishes
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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