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 = $\frac{1}{2}\left[\begin{array}{ccc}1& 1& 1\\ x_1& x_2& x_3\\ y_1& y_2& y_3\end{array}\right]$        (With co-ordinates of Vertices given)
                               A = $\frac{1}{2}\left[\begin{array}{ccc}1& 1& 1\\ a{t}^2& 0& a\\ 2at& at& 0\end{array}\right]$        
                              A = $\frac{1}{2}$[(0-a²t) + (2at.a - 0) + (at².at - 0)]
A = $\frac{1}{2}$ [ -at² + 2a²t + a²t³ ]
For A to be maximum: $\frac{d\mathrm{A}}{d\mathrm{t}}$ = 0
$\frac{d\mathrm{A}}{d\mathrm{t}}$ = $\frac{1}{2}$ [ -2at + 2a^​2 + 3a²t²]
$\frac{1}{2}$[ 3a²t² - 2at + 2a² ] = 0
3at² - 2t + 2a = 0
t = $\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(3\mathrm{a}\right)\left(2\mathrm{a}\right)}}{2\left(3\mathrm{a}\right)}$
 = $\frac{\left(2\right)\pm \sqrt{4-24{\mathrm{a}}^2}}{6a}$
= $\frac{1\pm \sqrt{1-6{\mathrm{a}}^2}}{3a}$
Therefore,
2at = $\frac{2\pm 2\sqrt{1-6{\mathrm{a}}^2}}{3}$
at² = a ${\left(\frac{1\pm \sqrt{1-6{\mathrm{a}}^2}}{3a}\right)}^2$ 
Hence Co-ordinates of A are (​a ${\left(\frac{1\pm \sqrt{1-6{\mathrm{a}}^2}}{3a}\right)}^2$ , $\frac{2\pm 2\sqrt{1-6{\mathrm{a}}^2}}{3}$ )
Regards