find the coordinates of the point A(t), 0<t<1 on the parabola y^2=4ax such that the area of triangle ABC (where C is the focus  and B is the point of intersection of tangents at A and vertex) is maximum.

Dear Student,
Vertex of parabola is (0 , 0)
Focus of parabola is (a , 0)
Co-ordinates of A are (at2 , 2at)
Co-ordinates of C are (a , 0)

Tangent at A is given by:
yt = x + at2 
Also tangent at vertex is the y - axis i.e. x = 0

For Co-ordinate of B put x = 0 in equation of tangent,
yt = 0 + at2 
y = at

Co-ordinates of B are (0 , at) 

Area of triangle ABC = 12111x1x2x3y1y2y3        (With co-ordinates of Vertices given)
                               A = 12111at20a2atat0        
                              A = 12[(0-a2t) + (2at.a - 0) + (at2.at - 0)]
A = 12 [ -at2 + 2a2t + a2t3 ]
For A to be maximum: dAdt = 0
dAdt12 [ -2at + 2a​2 + 3a2t2]
12[ 3a2t- 2at + 2a] = 0
3at- 2t + 2a = 0
t = --2±-22-43a2a23a
 = 2±4-24a26a
1±1-6a23a
Therefore,
2at = 2±21-6a23
at2 = a 1±1-6a23a2 
Hence Co-ordinates of A are (​a 1±1-6a23a2 , 2±21-6a23 )
Regards


 

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