The equation of the given curve is $x^2-4 x+4 y-8=0$. Match the following

$$ \begin{array}{lll} \hline & \text { List I } & \text { List II } \\ \hline \text { (A) } & \text { Focus } & \text { (I) }(4,2) \\ \hline \text { (B) } & \text { Vertex } & \text { (II) }(3,2) \\ \hline \text { (C) } & \begin{array}{l} \text { One end of the } \\ \text { latusrectum } \end{array} & \text { (III) }(2,3) \\ \hline \text { (D) } & \begin{array}{l} \text { point of intersection of the } \\ \text { axis and directrix } \end{array} & \text { (IV) }(2,4) \\ \hline & & \text { (V) }(2,2) \\ \hline \end{array} $$

$$ \text { The correct match is } $$

If one end of a focal chord of the parabola $y^2=\frac{8}{a} \times(a>0)$ is at $(1,4)$, then the length of this focal chord is

If the focal chord drawn through the point $(1,2)$ to the parabola $y^2=8 x$ meets this parabola in $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$, then $x_1+x_2=$

If $\left(2 t^2, 4 t\right)$ is a point on the parabola $y^2=8 x$ such that its focal distance is 3 , then $t=$