Write the parametric equations of the parabola: -
$3{x^2} + 3x + 7y + 8 = 0$

Hint: - For parabola equation ${X^2} = - 4aY$ the parametric coordinates are given as $\left( {2at, - a{t^2}} \right)$.

Complete step-by-step solution -
Given equation is $3{x^2} + 3x + 7y + 8 = 0$
First convert the equation into standard form
Divide by 3 in the given equation
${x^2} + x = - \dfrac{7}{3}y - \dfrac{8}{3}$
Now add and subtract by ${\left( {\dfrac{1}{2}} \right)^2}$in L.H.S to make a complete square in $x$.
$
  {x^2} + x + \dfrac{1}{4} - \dfrac{1}{4} = - \dfrac{7}{3}y - \dfrac{8}{3} \\
   \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = - \dfrac{7}{3}y - \dfrac{8}{3} + \dfrac{1}{4} \\
   \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = - \dfrac{7}{3}\left( {y + \dfrac{{29}}{{28}}} \right) \\
$
So, on comparing with standard equation of parabola which is ${X^2} = - 4aY$
$X = x + \dfrac{1}{2},{\text{ }}Y = y + \dfrac{{29}}{{28}},{\text{ }}4a = \dfrac{7}{3}$
As we know the parametric coordinates of the parabola ${X^2} = - 4aY$are$\left( {2at, - a{t^2}} \right)$.
Where $t$ is a parameter and this coordinates represents every point on the parabola.
Therefore parametric equation of the parabola is
$X = 2at$, and $Y = - a{t^2}$
$
   \Rightarrow x + \dfrac{1}{2} = 2.\dfrac{7}{{12}}t \\
   \Rightarrow x = - \dfrac{1}{2} + \dfrac{7}{6}t....................\left( 1 \right) \\
   \Rightarrow y + \dfrac{{29}}{{28}} = - \dfrac{7}{{12}}{t^2} \\
   \Rightarrow y = - \dfrac{{29}}{{28}} - \dfrac{7}{{12}}{t^2}................\left( 2 \right) \\
$
So, equation (1) and (2) represents the required parametric equation of the parabola.

Note: - In such types of questions first convert the equation into the standard form then compare it with standard equation of the parabola, then always remember the parametric coordinates of this parabola which is written above, then this parametric coordinates represents every point on the parabola so, satisfy these points and simplify we will get the required parametric equations of the parabola.