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  • Let a line pass through the point of intersection of the lines <img alt="\mathrm{b} x+10 y-8=0$ and $2 x-3 y=0, b \in \mathbf{R}-\left\

Let a line \mathrm{L} pass through the point of intersection of the lines \mathrm{b} x+10 y-8=0$ and $2 x-3 y=0, 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 \mathrm{17\left(x^{2}+y^{2}\right)=16}, then the eccentricity of the ellipse \mathrm{\frac{x^{2}}{5}+\frac{y^{2}}{\mathrm{~b}^{2}}=1} is :
 

Option: 1

\frac{2}{\sqrt{5}}


Option: 2

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


Option: 3

\frac{1}{\sqrt{5}}


Option: 4

\sqrt{\frac{2}{5}}


Answers (1)

best_answer

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

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Ritika Kankaria

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