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JEE MAIN - Mathematics (2022 - 29th July Morning Shift - No. 8)

Let a line L pass through the point of intersection of the lines $$b x+10 y-8=0$$ and $$2 x-3 y=0, \mathrm{~b} \in \mathbf{R}-\left\{\frac{4}{3}\right\}$$. If the line $$\mathrm{L}$$ 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 $$\frac{x^{2}}{5}+\frac{y^{2}}{\mathrm{~b}^{2}}=1$$ is :
$$ \frac{2}{\sqrt{5}} $$
$$\sqrt{\frac{3}{5}}$$
$$\frac{1}{\sqrt{5}}$$
$$\sqrt{\frac{2}{5}}$$

Explanation

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

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 $${x^2} + {y^2} = {{16} \over {17}}$$

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

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

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

$$ \Rightarrow 18{b^2} = 36$$

$$\therefore$$ $${b^2} = 2$$

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

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

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