Let a line L pass through the point of intersection of the lines bx...

Line is passing through intersection of $bx + 10y - 8 = 0 and 2x - 3y = 0$,

So its equation is given by,

\left(bx + 10y - 8\right) + \lambda\left(2x - 3y\right) = 0 . . . . . . . . . . \left(1\right)

And$the$line is passing through \left(1, 1\right) so$\lambda = b + 2$

Now putting the value of $\lambda in$equation \left(1\right) we get,

$⇒ \left(3b + 4\right)$x - $\left(3b - 4\right)$y - 8 = 0$</p><p>Now line \left(3b + 4\right)x - \left(3b - 4\right)y - 8 = 0$ is tangent to circle $17 \left({x}^{2} + {y}^{2}\right) = 16$

So $\dfrac{8}{\sqrt{{\left(3b + 4\right)}^{2} + {\left(3b - 4\right)}^{2}}} = \dfrac{4}{\sqrt{17}}$

$⇒ {b}^{2} = 2$

Now eccentricity of the ellipse $\dfrac{{x}^{2}}{5} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ will be,

${e}^{2} = 1 - \dfrac{{b}^{2}}{{a}^{2}}$

$⇒ e = \sqrt{1 - \dfrac{2}{5}} = \sqrt{\dfrac{3}{5}}$