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(B) The family of lines passing through the intersection of $bx + 10y - 8 = 0$ and $2x - 3y = 0$ is given by $(bx + 10y - 8) + \lambda(2x - 3y) = 0$.S

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The family of lines passing through the intersection of $bx + 10y - 8 = 0$ and $2x - 3y = 0$ is given by $(bx + 10y - 8) + \lambda(2x - 3y) = 0$.Since the line passes through $(1, 1)$,we have $(b + 10 - 8) + \lambda(2 - 3) = 0$,which gives $b + 2 - \lambda = 0$,so $\lambda = b + 2$.Substituting $\lambda$ back,the line equation is $(b + 2(2))x + (10 - 3(b + 2))y - 8 = 0$,which simplifies to $(b + 4)x + (4 - 3b)y - 8 = 0$.The line is tangent to the circle $x^2 + y^2 = \frac{16}{17}$,so the perpendicular distance from the origin $(0, 0)$ to the line equals the radius $r = \frac{4}{\sqrt{17}}$.Thus,$\frac{|-8|}{\sqrt{(b + 4)^2 + (4 - 3b)^2}} = \frac{4}{\sqrt{17}}$.Squaring both sides,$\frac{64}{b^2 + 8b + 16 + 16 - 24b + 9b^2} = \frac{16}{17} \implies \frac{4}{10b^2 - 16b + 32} = \frac{1}{17}$.$10b^2 - 16b + 32 = 68 \implies 10b^2 - 16b - 36 = 0 \implies 5b^2 - 8b - 18 = 0$.Wait,re-evaluating the intersection: $(b+2\lambda)x + (10-3\lambda)y - 8 = 0$. For $(1,1)$,$b+2\lambda + 10-3\lambda - 8 = 0 \implies b+2-\lambda = 0 \implies \lambda = b+2$.Line: $(b+2(b+2))x + (10-3(b+2))y - 8 = 0 \implies (3b+4)x + (4-3b)y - 8 = 0$.Distance: $\frac{8}{\sqrt{(3b+4)^2 + (4-3b)^2}} = \frac{4}{\sqrt{17}} \implies \frac{2}{\sqrt{9b^2+24b+16+16-24b+9b^2}} = \frac{1}{\sqrt{17}} \implies \frac{4}{18b^2+32} = \frac{1}{17} \implies 18b^2+32 = 68 \implies 18b^2 = 36 \implies b^2 = 2$.For the ellipse $\frac{x^2}{5} + \frac{y^2}{2} = 1$,$a^2 = 5$ and $b^2 = 2$. Since $a^2 > b^2$,$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2}{5}} = \sqrt{\frac{3}{5}}$.

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