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Analytical Geometry
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ankur singh
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Joined: 24 Feb 2007
Posts: 3
24 Feb 2007 15:17:12 IST
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ghj
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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 }
PTS =
;{ where Tan
= 1 / t }Let the angle bt tangents be
;
;So, ext angle
PSG = {
+(
/2 -
)}
PSG = {
+(
/2 -
)} Just expand..Tan
PSG { = 2t / (t2-1) } ;
PSG { = 2t / (t2-1) } ;Solve for cot
;{u get it as 1/t }
;{u get it as 1/t }So,
= Tan-1(1/t) ....ans.
= Tan-1(1/t) ....ans.
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
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












