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 $m$ is the length of the latusrectum and $n$ is the length of the major-axis of the ellipse $25 x^2+16 y^2-150 x-64 y-111=0$, then the ordered pair $(m, n)=$

If $P(\theta)$ and $Q\left(\frac{\pi}{2}+\theta\right)$ are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the locus of mid-point of $P Q$ is $\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1$, then $\frac{a+b}{\alpha+\beta}=$