Let the area of the triangle with vertices $A(1,\alpha ), B(\alpha ,0)$ and $C(0,\alpha )$ be 4 sq. units. If the points $(\alpha ,-\alpha ),(-\alpha ,\alpha )$ and $\left({\alpha }^2,\beta \right)$ are collinear, then $\beta$ is equal to

$$\begin{gathered} \frac{1}{2}\left|\begin{array}{l}\alpha \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|=\pm 4 \\ \alpha \left(0\right)-0+1\left(\alpha \right)=\pm 8 \end{gathered}$$

$\Rightarrow \alpha =\pm 8$

$\text{ Now given points }(8,-8),(-8,8),(64,\beta )$

$$\begin{gathered} A(-8,8),B(8,-8),C (64,\beta ) are collinear \\ \therefore slope of AB=Slope of AC \\ \frac{-8-8}{8+8}=\frac{\beta -8}{64+8}\Rightarrow \beta =-64 \end{gathered}$$