A. Equation of tangent is $y y_1-2\left(x+x_1\right)=0$
So, $\quad \sqrt{8} y=2(x+2 \Rightarrow \sqrt{8} y=2 x+4$
$$
\begin{aligned}
\Rightarrow & 2 \sqrt{2} y & =2 x+4 \\
\Rightarrow & -\sqrt{2} y+x+2 & =0
\end{aligned}
$$
B. Equation of normal is $y=m x-2 a m-a m^3$
Here, $y^2=16 x$
$$
\begin{array}{ll}
\therefore & a=4 \\
\text { and } & m=\tan 45^{\circ}=1 \\
\text { So, } & y=x-8-4 \Rightarrow x-y-12=0
\end{array}
$$
C.

Given, $y^2=12 x$;
$\therefore \quad 4 a=12 \Rightarrow a=3$
For focal chord $t_1 t_2=-1 \quad \Rightarrow t_2=-\frac{1}{t_1}$
$\therefore \quad\left(x_1, y_1\right)=\left(3 t^2, 6 t\right)$ and $\left(x_2, y_2\right)=\left(\frac{3}{t^2},-\frac{6}{t}\right)$
Now, $\quad y_1 y_2=6 t \times\left(-\frac{6}{t}\right)=-36$
D. We have, $\quad y^2=k x-16 \Rightarrow y^2=k\left(x-\frac{16}{k}\right)$
On comparing it with $y^2=4 a x$, we get
$\therefore \quad 4 a=k \quad \Rightarrow a=\frac{k}{4}$


Again its directrix is given by
$$
\Rightarrow \quad \begin{aligned}
x-\frac{16}{k} & =-a \\
x \quad x & =\frac{16}{k}-\frac{k}{4}
\end{aligned}
$$
On comparing it with $x-3=0$, we get
$$
\begin{aligned}
& \frac{16}{k}-\frac{k}{4}=3 \\
& \Rightarrow \quad 64-k^2=12 k \\
& \Rightarrow \quad k^2+12 k-64=0 \\
& \Rightarrow \quad k=4,-16 \\
& \Rightarrow \quad k=4 \quad[k \neq-16] \\
& \text { (a) } \rightarrow \text { (iv); (b) } \rightarrow \text { (vi); (c) } \rightarrow \text { (i); (d) } \rightarrow \text { (ii) } \\
&
\end{aligned}
$$