QJEEM22Mad822: JEE Main 2022 Mathematics - Class 11 JEEM22Mad822 Solution: Let a line L pass through the point of intersection of the…

${L_1}:bx + 10y - 8 = 0,\,{L_2}:2x - 3y = 0$

then $L:(bx + 10y - 8) + \lambda (2x - 3y) = 0$

$\because$ It passes through $(1,\,1)$

$\therefore$ $b + 2 - \lambda = 0 \Rightarrow \lambda = b + 2$

and touches the circle ${x^2} + {y^2} = {{16} \over {17}}$

$\left| {{{{8^2}} \over {{{(2\lambda + b)}^2} + {{(10 - 3\lambda )}^2}}}} \right| = {{16} \over {17}}$

$\Rightarrow 4{\lambda ^2} + {b^2} + 4b\lambda + 100 + 9{\lambda ^2} - 60\lambda = 68$

$\Rightarrow 13{(b + 2)^2} + {b^2} + 4b(b + 2) - 60(b + 2) + 32 = 0$

$\Rightarrow 18{b^2} = 36$

$\therefore$ ${b^2} = 2$

$\therefore$ Eccentricity of ellipse : ${{{x^2}} \over 5} + {{{y^2}} \over {{b^2}}} = 1$ is

$\therefore$ $e = \sqrt {1 - {2 \over 5}} = \sqrt {{3 \over 5}}$