If the maximum distance of normal to the ellipse $$\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1, b < 2$$, from the origin is 1, then the eccentricity of the ellipse is :

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 :

The acute angle between the pair of tangents drawn to the ellipse $$2 x^{2}+3 y^{2}=5$$ from the point $$(1,3)$$ is :

If the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ meets the line $$\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$$ on the $$x$$-axis and the line $$\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$$ on the $$y$$-axis, then the eccentricity of the ellipse is :