A) $$\begin{gathered} Given parabola 4y^2-32x+4y+65=0 \\ \Rightarrow 4(y^2+y)=32x-65 \\ \Rightarrow {(y+\frac{1}{2})}^2=8(x-2) \\ \therefore \mathrm{a}=2 and h=2 \\ \because \mathrm{x}-\mathrm{h}>2\mathrm{a} \\ \Rightarrow \mathrm{x}-2>4 \\ \Rightarrow \mathrm{x}>6 \end{gathered}$$

B)

 $$\begin{gathered} \frac{(\mathrm{x}+1{)}^2}{9}+\frac{(\mathrm{y}-2{)}^2}{4}=1 where \mathrm{a}=3,b=2 \\ centre=(-1,2) and given point=(1,2) \\ Now \text{ minimum distance (L)}=3-dis\tan ce between (-1,2) and (1,2) \\ =3-2=1 \end{gathered}$$

         Maximum  distance(G) = 3+ distance between (–1,2)  and (1,2)=3+2=5

Now G-L=4

C) $12(\mathrm{x}-1{)}^2-4(\mathrm{y}-4{)}^2=75,$

      $\begin{array}{r}\frac{(\mathrm{x}-1{)}^2}{\left(\frac{75}{12}\right)}-\frac{(\mathrm{y}-4{)}^2}{\frac{75}{4}}=1, \mathrm{e}=2 \\ \therefore \text{ Foci =}(\mathrm{h}\pm \mathrm{ae}, \mathrm{k})=(6, 4) and (-4,4) \\ By comparision k=4 \\ \end{array}$