Question

Let a line $L$ pass through the point of intersection of the lines $bx+10y-8=0$ 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 $17\left(x^2+y^2\right)=16$, then the eccentricity of the ellipse $\frac{x^2}{5}+\frac{y^2}{b^2}=1$ is :

• A. $\frac{2}{\sqrt{5}}$
• B. $\sqrt{\frac{3}{5}}$
• C. $\frac{1}{\sqrt{5}}$
• D. $\sqrt{\frac{2}{5}}$

Answer 1

Answer: (B)

Line is passing through intersection of $bx+10y-8=0$ and $2x-3y=0$ is $(bx+10y-8)+\lambda (2x-3y)=0$. As line is passing through $(1,1)$ so $\lambda =b+2$
Now line $(3b+4)x-(3b-4)y-8=0$ is tangent to circle $17\left(x^2+y^2\right)=16$
$\text{ So }\frac{8}{\sqrt{(3b+4{)}^2+(3b-4{)}^2}}=\frac{4}{\sqrt{17}}$
$\Rightarrow b^2=2\Rightarrow e=\sqrt{\frac{3}{5}}$