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

Question

Let a line $L$ pass through the point of intersection of the lines $bx+10y-8=0$ and $2x-3y=0$, $b\in\mathbb R-\left\{\frac43\right\}$. 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 $\dfrac{x^2}{10}+\dfrac{y^2}{b^2}=1$ is:

Accepted Answer

$L:(bx+10y-8)+\lambda(2x-3y)=0$; through $(1,1)$: $(b+2)-\lambda=0\Rightarrow\lambda=b+2$.

$L:(3b+4)x+(4-3b)y-8=0.$

Tangent to $x^2+y^2=\frac{16}{17}$: distance from origin equals radius:
$$\frac{64}{(3b+4)^2+(4-3b)^2}=\frac{16}{17}\ \Longrightarrow\ 18b^2+32=68\ \Longrightarrow\ b^2=2.$$

With $a^2=10$: $e=\sqrt{1-\dfrac{b^2}{a^2}}=\sqrt{1-\dfrac2{10}}=\sqrt{\dfrac45}=\dfrac2{\sqrt5}.$

Solutions · 1