If the $x$ -intercept of a focal chord of the parabola $y^2=8x+4y+4\text{ is }3$, then the length of this chord is equal to _______

Given ${\mathrm{y}}^2=8\mathrm{x}+4\mathrm{y}+4$
$$\begin{gathered} \begin{array}{l}\Rightarrow {\mathrm{y}}^2-4\mathrm{y}=8\mathrm{x}+4 \\ \Rightarrow {\mathrm{y}}^2-4\mathrm{y}+4=8\mathrm{x}+8 \\ \Rightarrow (\mathrm{y}-2{)}^2=8(\mathrm{x}+1)\end{array} \end{gathered}$$
Parabola with vertex V(-1, 2) and $4a=8$   $\Rightarrow \mathrm{a}=2$
So focus $\mathrm{S}(\mathrm{a}+\mathrm{\alpha },\mathrm{\beta })=\mathrm{S}(2-1,2)=(1,2)$
Let the equation of focal chord of the parabola $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$and given x-intercept of focal chord is ‘3’. 
so $\frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{\mathrm{b}}=1$and passing through focus (1, 2)
so $\frac{1}{3}+\frac{2}{\mathrm{b}}=1$
$\frac{2}{\mathrm{b}}=1-\frac{1}{3}=\frac{2}{3}\Rightarrow \mathrm{b}=3$
Equation of focal chord $\mathrm{x}+\mathrm{y}=3$
So, slope of focal chord
$\begin{array}{l}\therefore \mathrm{m}=-1\\ \tan \mathrm{\theta }=-1\Rightarrow \mathrm{\theta }=\frac{3\mathrm{\pi }}{4}\end{array}$
We know that length of focal chord in$4{\mathrm{acosec}}^2\mathrm{\theta }=4\left(2\right)(\sqrt{2}{)}^2$ $=8\left(2\right)=16$