Find the coordinates of the vertex, focus, and the equations of the directrix and axes of the parabolas
$$\begin{gathered} \mathrm{i}) {\mathrm{y}}^2-\mathrm{x}+4\mathrm{y}+5=0 \\ \text{ ii) }3{\mathrm{x}}^2-9\mathrm{x}+5\mathrm{y}-2=0 \\ \text{ iii) }{\mathrm{x}}^2+8\mathrm{x}+12\mathrm{y}+4=0 \end{gathered}$$

$\text{ i) Given parabola }{\mathrm{y}}^2-\mathrm{x}+4\mathrm{y}+5=0$
$\begin{array}{l}{\mathrm{y}}^2+4\mathrm{y}+4-4=\mathrm{x}-5\\ (\mathrm{y}+2{)}^2=\mathrm{x}-5+4\Rightarrow (\mathrm{y}+2{)}^2=1(\mathrm{x}-1)\end{array}$
Compare $(\mathrm{y}-\mathrm{k}{)}^2=4\mathrm{a}(\mathrm{x}-\mathrm{h})$
$4\mathrm{a}=1\Rightarrow \mathrm{a}=\frac{1}{4}$
$\text{ Vertex }\mathrm{A}=(\mathrm{h},\mathrm{k})=(1,-2)$
$\text{ Focus }\mathrm{S}=(\mathrm{h}+\mathrm{a},\mathrm{k})=\left(1+\frac{1}{4},-2\right)=\left(\frac{5}{4},-2\right)$
Equation of directrix of x - h + a = 0
$\mathrm{x}-1+\frac{1}{4}=0\Rightarrow 4\mathrm{x}-4+1=0\Rightarrow 4\mathrm{x}-3=0$
$\text{ Equation of axis is }\mathrm{y}-\mathrm{k}=0\Rightarrow \mathrm{y}+2=0$
$\text{ ii) Given parabola is }3{\mathrm{x}}^2-9\mathrm{x}+5\mathrm{y}-2=0$
$\begin{array}{l}3\left({\mathrm{x}}^2-3\mathrm{x}\right)=-5\mathrm{y}+2\\ 3{\left(\mathrm{x}-\frac{3}{2}\right)}^2=-5\mathrm{y}+2+\frac{27}{4}=-5\left(\mathrm{y}-\frac{7}{4}\right)\\ {\left(\mathrm{x}-\frac{3}{2}\right)}^2=-\frac{5}{3}\left(\mathrm{y}-\frac{7}{4}\right)\end{array}$
$\text{ Compare with }(\mathrm{x}-\mathrm{h}{)}^2=-4\mathrm{a}(\mathrm{y}-\mathrm{k})$
$4\mathrm{a}=\frac{5}{3}\Rightarrow \mathrm{a}=\frac{5}{12}$
$$\begin{gathered} \text{ Vertex }\mathrm{A}=(\mathrm{h},\mathrm{k})=\left(\frac{3}{2},\frac{7}{4}\right) \\ \text{ Focus }\mathrm{S}=(\mathrm{h},\mathrm{k}-\mathrm{a})=\left(\frac{3}{2},\frac{7}{4}-\frac{5}{12}\right)=\left(\frac{3}{2},\frac{4}{3}\right) \\ \text{ Equation of directrix is }\mathrm{y}-\mathrm{k}-\mathrm{a}=0 \\ \mathrm{y}-\frac{7}{4}-\frac{5}{12}=0\Rightarrow 6\mathrm{y}=13 \\ \text{ Equation of axis is }\mathrm{x}-\mathrm{h}=0 \\ \mathrm{x}-\frac{3}{2}=0\Rightarrow 2\mathrm{x}-3=0 \end{gathered}$$
$$\begin{gathered} \text{ iii) Given parabola is }{\mathrm{x}}^2+8\mathrm{x}+12\mathrm{y}+4=0 \\ \begin{array}{l}{\mathrm{x}}^2+8\mathrm{x}+16-16+12\mathrm{y}+4=0\\ (\mathrm{x}+4{)}^2=-12\mathrm{y}+12=-12(\mathrm{y}-1\left) \\ \text{ Compare with }\right(\mathrm{x}-\mathrm{h}{)}^2=-4\mathrm{a}(\mathrm{y}-\mathrm{k}) \\ 4\mathrm{a}=12\Rightarrow \mathrm{a}=3\end{array} \end{gathered}$$
$$\begin{gathered} \text{ Vertex }\mathrm{A}=(\mathrm{h},\mathrm{k})=(-4,1) \\ \text{ Focus }\mathrm{S}=(\mathrm{h},\mathrm{k}-\mathrm{a})=(-4,1-3)=(-4,-2) \\ \text{ Equation of directrix is }\mathrm{y}-\mathrm{k}-\mathrm{a}=0 \\ \mathrm{y}-1-3=0\Rightarrow \mathrm{y}-4=0 \\ \text{ Equation of axis is }\mathrm{x}-\mathrm{h}=0\Rightarrow \mathrm{x}+4=0 \end{gathered}$$