Question

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

Answer 1

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

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

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

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

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

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

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

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

$\Rightarrow 18b^2=36\therefore b^2=2$

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

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