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

<p>${L_1}:bx + 10y - 8 = 0,\,{L_2}:2x - 3y = 0$</p> <p>then $L:(bx + 10y - 8) + \lambda (2x - 3y) = 0$</p> <p>$\because$ It passes through $(1,\,1)$</p> <p>$\therefore$ $b + 2 - \lambda = 0 \Rightarrow \lambda = b + 2$</p> <p>and touches the circle ${x^2} + {y^2} = {{16} \over {17}}$</p> <p>$$\left| {{{{8^2}} \over {{{(2\lambda + b)}^2} + {{(10 - 3\lambda )}^2}}}} \right| = {{16} \over {17}}$$</p> <p>$$ \Rightarrow 4{\lambda ^2} + {b^2} + 4b\lambda + 100 + 9{\lambda ^2} - 60\lambda = 68$$</p> <p>$\Rightarrow 13{(b + 2)^2} + {b^2} + 4b(b + 2) - 60(b + 2) + 32 = 0$</p> <p>$\Rightarrow 18{b^2} = 36$</p> <p>$\therefore$ ${b^2} = 2$</p> <p>$\therefore$ Eccentricity of ellipse : ${{{x^2}} \over 5} + {{{y^2}} \over {{b^2}}} = 1$ is</p> <p>$\therefore$ $e = \sqrt {1 - {2 \over 5}} = \sqrt {{3 \over 5}}$</p>