Question

Let a line L pass through the point intersection of the lines $bx \+ 10y – 8 = 0$ and 2x - 3y = 0, \quad b \in \mathbb{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 $\frac{x^2}{5} + \frac{y^2}{5} = 1$ is

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

Answer 1

Answer: (B)

$\frac{\sqrt{3}}{5}$

Answer 2

$L_1 :bx \+ 10y – 8 = 0, L_2 : 2x – 3y = 0$
then $L : (bx \+ 10y – 8) + λ(2x – 3y) = 0$
Since, It passes through $(1, 1)$
so, $b \+ 2 – λ = 0 ⇒ λ = b \+ 2$
and touches the circle $x^2 + y^2 = \frac{16}{17}$

$\left|\frac{82}{(2\lambda + b)^2} + (10 - 3\lambda)^2\right| = \frac{16}{17}$

$⇒$ $4\lambda^2 + b^2 + 4b\lambda + 100 + 9\lambda^2 - 60\lambda = 68$

$⇒$ $13(b+2)^2 + b^2 + 4b(b+2) - 60(b+2) + 32 = 0$
⇒$$18b^2=36 ∴b^2=2

∴ Eccentricity of ellipse $\frac{x^2}{5} + \frac{y^2}{b^2} = 1$ is
$e = \sqrt{1 - \frac{2}{5}}$
$e = \frac{\sqrt{3}}{5}$
So, the correct option is (B): $\frac{\sqrt{3}}{5}$