If F₁ and F₂ are the feet of the perpendiculars from the foci S₁ and S₂ of the ellipse $\left({\mathrm{x}}^2/25\right)+\left({\mathrm{y}}^2/16\right)=1$ on the tangent at any point P on the ellipse, then the least value of $\frac{{\mathrm{S}}_1{\mathrm{F}}_1+{\mathrm{S}}_2{\mathrm{F}}_2}{3}=.....$

Equation of the ellipse is  $\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{16}=1$

$$\begin{gathered} where a^2=25,b^2=16 \\ We know that product of perpendiculars from foci to any \tan gent is b^2 \\ \Rightarrow \left(S_1{F}_1\right)\left(S_2{F}_2\right)=16 \\ Since A.M\ge G.M \\ \Rightarrow S_1{F}_1+S_2{F}_2\ge 2\sqrt{\left(S_1{F}_1{S}_2{F}_2\right)} \\ \Rightarrow S_1{F}_1+S_2{F}_2\ge 2\left(4\right)=8 \\ minimum value of S_1{F}_1+S_2{F}_2=8 \\ Now \frac{S_1{F}_1+S_2{F}_2}{3}=\frac{8}{3}=2.6 \end{gathered}$$
$\Rightarrow \text{ Least value of }\frac{{\mathrm{S}}_1{\mathrm{F}}_1+{\mathrm{S}}_2{\mathrm{F}}_2}{3}=\frac{2\mathrm{b}}{3}=\frac{8}{3}=2.666$