If the point
[tex](a \: \: a \sqrt{3} )[/tex]
lies on
[tex]y = 2x[/tex]
then it must satisfy it.
That is
[tex]a + \sqrt{3} = 2(a)[/tex]
[tex] \rightarrow \: a + \sqrt{3} = 2a[/tex]
Grouping like terms we have,
[tex] \sqrt{3} = 2a - a[/tex]
[tex] \therefore \: a = \sqrt{3} [/tex]