A curve is given parametrically by the equations $ x = 3qt^2$ , $y = 4( t^3 + 1 )$ , where $q$ is a constant. The curve meets the x-axis at $X$ and the y-axis at $Y$. Given that $OX = 2OY$, where $O$ is the origin, find the value of $q$.
So, I've worked out that the point X is $(3q,0)$ and Y is $(0,4)$
However, I'm unsure what the notation $OX = 2OY$ means.
$OX = 2OY$ means the length of $OX $ is double the length of $OY $.
Given that $X = (3q, 0) $, we have $OX = |3q|$. Also $Y = (0, 4) \Rightarrow OY = 4$. Putting it all together yields $|3q| = 8$.