Let a line L pass through the point of intersection of the lines b x+10 y-8=0 and [2 x-3 y…

Question

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 :

Accepted Answer

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

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

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

Solutions · 1