Question
A curve is given by parametric equations $x=qt+2$ and $y=t^3 + q$, where $q$ is a constant. The $y$-intercept of the curve is at $\left ( 0,\frac{617}{25} \right )$. Find the value(s) of $q$ and hence state the parametric equations.
Solution
$x=qt+2$ $\hspace{2.6cm}$ [1]
$y=t^3 + q$ $\hspace{2.7cm}$ [2]
From [1],
$\Rightarrow t=\frac{x-2}{q}$ $\hspace{2.5cm}$ [3]
Substituting [3] into [2] gives
$\ y=\left ( \frac{x-2}{q} \right )^3 +q$ $\hspace{1.4cm}$ [4]
The curve crosses the $y$-axis when $x=0$, so substituting $x=0$ and $y=\frac{617}{25}$ into [4] gives
$\frac{617}{25}=\frac{-8}{q^3}+q$
I want to know what is the easiest way to solve this equation?