(A)        Q Light rays after reflection passes through focus and reflected ray is parallel to directrix.

$\therefore$         point (1, 2) will be one extremity of LR.

$\therefore$        Length of LR

$=2\sqrt{{(2-1)}^2+{(2+\sqrt{3}-2)}^2}=4.$

(B)        Q Locus of centres of circles will be tangent to hyperbola at (4, 3)

            i.e.  = 1 ⇒ 3x – 2y = 6

$\therefore$        Area bounded =  × 2 × 3 = 3.

(C)     $\because$ Tangent at upper‑end (3, 6) will be y · 6 = 6 (x + 3) ⇒ x – y = – 3         ….…(1)

            Equation of C.O.C. for (– 2α, α) w.r.t. ellipse is $\frac{\mathrm{x}(-2\mathrm{\alpha })}{{\mathrm{a}}^2}+\frac{\mathrm{y\alpha }}{{\mathrm{b}}^2}=1$                           ….…(2)

       $\because$  (1) and (2) represent same line

         $\therefore \frac{{\mathrm{a}}^2}{-2\mathrm{\alpha }}=\frac{-{\mathrm{b}}^2}{\mathrm{\alpha }}=\frac{-3}{1}$

            ⇒ a² = 6α and b² = 3α

    $\therefore \frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}=2.$

(D)    $\because \mathrm{f}\left(\mathrm{\alpha }\right)=\sqrt{{\mathrm{\alpha }}^2+9}+\sqrt{{(\mathrm{\alpha }-7)}^2+16}$ is equivalent of sum of distances of (α, 0) from (0, 3) and (7, 4). 

For minimum value, (α, 0) will be point of intersection of line joining (7, 4) and (0, – 3) with x‑axis.

$\therefore$   α = 3