Convexity of the set $S=\{(x_1,x_2):x_2^2\geq8x_1\}$ 
Using Algebraic approach, test the convexity of the set $$S=\{(x_1,x_2):x_2^2\geq8x_1\}$$

Definition of convexity: $S \in \mathbb R^2$ is a convex set if $\forall \alpha \in \mathbb R, 0 \leq\alpha \leq 1$ and $\forall \vec x,\vec y \in S$ holds: $\alpha \vec x + (1 - \alpha)\vec y \in S$.
Let $\vec x=(a,b)$ and $\vec y=(c,d)$ then the convex combination is $\alpha(a,b)+(1-\alpha)(c,d)=(\alpha(a-c)+c,\alpha(b-d)+d)$ and
$$
\begin{align}
b^2-8a&\geq0\\
d^2-8c&\geq0\qquad(A)
\end{align}
$$
Now I need to show that,
$$
\begin{align}
&(\alpha(b-d)+d)^2-8(\alpha(a-c)+c)\\
&\alpha^2b^2-2\alpha^2bd+\alpha^2d^2 +d^2+2\alpha(b-d)d-\alpha(8a)+\alpha(8c)-8c\\
&\alpha^2b^2-2\alpha^2bd+\alpha^2d^2+(d^2-8c)+2\alpha(b-d)d-\alpha(8a)+\alpha(8c) 
\end{align}
$$
I couldn't conclude anything from the above expression except for $d^2-8c\geq0$. Any help will be appreciated.
 A: There is a mistake in your expansion. The convex combination is indeed
$$\vec x_\alpha = (\alpha(a-c)+c, \alpha(b-d)+d),$$
but when you squared the value $\alpha(b-d)+d$, you should have gotten
$$\alpha^2(b-d)^2+d^2+2\alpha(b-d)d = \alpha^2b^2-2\alpha^2bd+\alpha^2d^2 +d^2+2\alpha(b-d)d.$$
Subtractig $8(\alpha(a-c)+c)$ from that, you the result is not what you have, as the result should have, somewhere, some $\alpha^2b^2$ term.
A: We have $(\pm 2\sqrt2)^2=8 \ge 8(1)$ hence we can conclude that
$(1, 2\sqrt2)\in S$ and $(1, -2\sqrt2)\in S$.
However, their midpoint, $(1, 0)$ is not in $S$ since
$$0^2 < 8(1)$$
It is not convex.

We can make it more general, let $t>0$, then we have $(t, \pm 2\sqrt2 t) \in S$ but $(t,0)\notin S.$

More detailed reasoning in proving non-convexity.
If $S$ is convex, then for any $(a, b), (c,d) \in S$, then $\forall \alpha \in (0,1),  \alpha (a, b) + (1-\alpha)(c,d) \in S. $
To exhibit that $S$ is not convex, it means that we can find $(a,b), (c,d) \in S$ such that there is such an $\alpha \in (0,1)$, $\alpha (a,b) + (1-\alpha) (c,d) \in S$.
We pick $(a,b)=(1, 2\sqrt2), (c,d) = (1, -2\sqrt2)$, we can check that $(a,b), (c,d) \in S$.  Pikcing $\alpha =0.5$ illustrates that it is not convex.
