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Find the area of $\Delta ABC$ where AC is focal chord of the parabola $y^2=4ax$ & B is the vertex of parabola $y^2=4ax$?

$\begin{array}{1 1}(a)\;a^2|t_1t_2(t_1-t_2)|\\(b)\;a|t_1t_2(t_1t_2)|\\(c)\;2a^2|t_1t_2(t_1t_2)|\\(d)\;a^2|t_1t_2(t_1+t_2)|\end{array}$

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A)
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Vertex B(0,0)
For $\Delta ABC$,$A(at_1^2,2at_1),B(0,0),C(at_2^2,2at_2)$.
Area of $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$
Where $s=\large\frac{a+b+c}{2}$
$a$- distance between AB.
$b$- distance between BC
$c$- distance between AC.
$AB=\sqrt{(at_1^2)^2+(2at_1)^2}=a$
$BC=\sqrt{(at_2^2)^2+(2at_2)^2}=b$
$AC=\sqrt{a(t_2-t_1^2)^2+(2a(t_2-t_1))^2}=c$
Area of $\Delta ABC=\large\frac{1}{2}$$\begin{vmatrix}at_1^2&2at_1&1\\at_2^2&2at_2&1\\0&0&1\end{vmatrix}$
$\Rightarrow \large\frac{1}{2}$$\mid 2a^2t_1^2t_2-2a^2t_1t_2^2\mid$
$\Rightarrow \large\frac{1}{2}$$\times 2\mid a^2t_1^2t_2-a^2t_1t_2^2\mid$
$\Rightarrow a^2\mid t_1t_2-t_1t_2^2\mid$
$\Rightarrow a^2\mid t_1t_2(t_1-t_2)\mid$
Hence (a) is the correct answer.
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