Given: $A\left(1,2\right), B\left(4,-4\right), C\left(2,2\sqrt{2}\right)$ be points on the parabola $y^2=4x·$$\alpha$ and $\beta$, respectively represent the area of $\mathrm{\Delta }ABC$ and the area of the triangle formed by the tangents at $A\mathit{,}\mathit{ }B\mathit{,}\mathit{ }C$ to the above parabola

Area of $△ABC=\frac{1}{2}\left|\begin{array}{ccc}1& 2& 1\\ 4& -4& 1\\ 2& 2\sqrt{2}& 1\end{array}\right|=3\sqrt{2}$

So $\alpha =3\sqrt{2}$

We know that the area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points

$\Rightarrow \alpha =2\beta$

So $\beta =\frac{3\sqrt{2}}{2}$

$\Rightarrow \alpha \beta =9$