# 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]
$\frac{1}{2}$[ 3a2t- 2at + 2a] = 0
3at- 2t + 2a = 0
t = $\frac{-\left(-2\right)±\sqrt{{\left(-2\right)}^{2}-4\left(3\mathrm{a}\right)\left(2\mathrm{a}\right)}}{2\left(3\mathrm{a}\right)}$
= $\frac{\left(2\right)±\sqrt{4-24{\mathrm{a}}^{2}}}{6a}$
$\frac{1±\sqrt{1-6{\mathrm{a}}^{2}}}{3a}$
Therefore,
2at = $\frac{2±2\sqrt{1-6{\mathrm{a}}^{2}}}{3}$
at2 = a ${\left(\frac{1±\sqrt{1-6{\mathrm{a}}^{2}}}{3a}\right)}^{2}$
Hence Co-ordinates of A are (​a ${\left(\frac{1±\sqrt{1-6{\mathrm{a}}^{2}}}{3a}\right)}^{2}$ , $\frac{2±2\sqrt{1-6{\mathrm{a}}^{2}}}{3}$ )
Regards

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