Let a line L pass through the point of intersection of the lines $\mathrm{bx}+10\mathrm{y}-8=0\text{ and }2\mathrm{x}-3\mathrm{y}=0,\mathrm{b}\in \mathrm{R}-\left\{\frac{4}{3}\right\}$. If the line L also passes through the point (1,1) and touches the circle $17\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)=16$, then the eccentricity of the ellipse $\frac{{\mathrm{x}}^2}{5}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$is:

Line L is given by $(\mathrm{bx}+10\mathrm{y}-8)+\mathrm{\lambda }(2\mathrm{x}-3\mathrm{y})=0$ 
$\Rightarrow (\mathrm{b}+2\mathrm{\lambda })\mathrm{x}+(10-3\mathrm{\lambda })\mathrm{y}-8=0$
As line L is passing through (1,1) is  $\mathrm{b}+2\mathrm{\lambda }+10-3\mathrm{\lambda }-8=0$ $\Rightarrow \mathrm{b}+2=\mathrm{\lambda }$
Now, L becomes $(\mathrm{b}+2(\mathrm{b}+2\left)\right)\mathrm{x}+(10-3(\mathrm{b}+2\left)\right)\mathrm{y}-8=0$
$\Rightarrow (3\mathrm{b}+4)\mathrm{x}-(3\mathrm{b}-4)\mathrm{y}-8=0$
Now, L is also a tangent to circle $17\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)=16$
$\begin{array}{l}\therefore \frac{|0-0-8|}{\sqrt{(3\mathrm{b}+4{)}^2+(3\mathrm{b}-4{)}^2}}=\sqrt{\frac{16}{17}}\\ \Rightarrow \frac{64}{9{\mathrm{b}}^2+16+24\mathrm{b}+9{\mathrm{b}}^2+16-24\mathrm{b}}=\frac{16}{17}\\ \Rightarrow 18{\mathrm{b}}^2+32=68\Rightarrow 18{\mathrm{b}}^2=36\Rightarrow {\mathrm{b}}^2=2\end{array}$
Now eccentricity of the ellipse $\frac{{\mathrm{x}}^2}{5}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\text{ is }\mathrm{e}=\sqrt{1-\frac{2}{5}}$
$\Rightarrow \mathrm{e}=\sqrt{\frac{3}{5}}$