$\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{3}=1$

Equation of PQ : x – h = 0

If R (a, b) then chord of contact of R is PQ

i.e.        C.O.C. = T = 0

 $\Rightarrow \frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{3}=1$                       .....(1)

and also    $\frac{\mathrm{x}}{\mathrm{h}}$  = 1           .....(2)

Comparing (1) and (2)

 $\frac{\frac{\mathrm{\alpha }}{4}}{\frac{1}{\mathrm{h}}}=\frac{\frac{8}{3}}{0}=\frac{1}{1}$

 $\Rightarrow \mathrm{\alpha }=\frac{4}{\mathrm{h}};\mathrm{\beta }=0\Rightarrow \mathrm{Point}\mathrm{R}\mathrm{is}(\frac{4}{\mathrm{h}},0)$

coordinate of P and Q are respectively  $\mathrm{P}({\mathrm{h}}_1+\sqrt{{(1+\frac{{\mathrm{h}}^2}{4})}^3})\mathrm{and}\mathrm{Q}({\mathrm{h}}_1-\sqrt{{(1-\frac{{\mathrm{h}}^2}{4})}^3})$

area  $\left(\mathrm{\Delta PQR}\right)=\frac{1}{2}\left(\mathrm{PQ}\right)\left(\mathrm{MR}\right)=\frac{1}{2}\times 2\sqrt{{(1-\frac{{\mathrm{h}}^2}{4})}^3}\times (\frac{4}{\mathrm{h}}-\mathrm{h})$

 $\mathrm{\Delta }\left(\mathrm{h}\right)=\frac{{(4-{\mathrm{h}}^2)}^{3/2}\sqrt{3}}{2\mathrm{h}}$

Since  Δ(h) is decreasing

 ${\text{∴Δ}}_1=\mathrm{\Delta }{\left(\mathrm{h}\right)}_{\max .}=\mathrm{\Delta }\mathrm{\hspace{0.17em}}\left(\frac{1}{2}\right)=\frac{45}{8}\sqrt{5}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{\Delta }}_2=\mathrm{\Delta }{\left(\mathrm{h}\right)}_{\min .}=\mathrm{\Delta }\left(1\right)=\frac{9}{2}$

Hence,  $\frac{8}{\sqrt{5}}{\mathrm{\Delta }}_1-8{\mathrm{\Delta }}_2=9$.