Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P of n sides and q be the number of all quadrilaterals that can be formed by joining the vertices of P. If p + q = 126, then the eccentricity of the ellipse \frac{x^2}{16} + \frac{y^2}{n} = 1 is :

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Answer: A. \frac{1}{\sqrt{2}} \begin{aligned} & 	ext { Total trangles }=\Rightarrow={ }^{\mathrm{h}} \mathrm{C}_3 \ & 	ext { Total auadrilaterals }={ }^{\mathrm{h}} \mathrm{C}_4=\mathrm{q} \ & { }^{\mathrm{n}} \mathrm{C}_3+{ }^{\mathrm{n}} \mathrm{C}_4=126 \Rightarrow{ }^{\mathrm{n}+1} \mathrm{C}_4=126 \ & \Rightarrow \mathrm{n}+1=9 \Rightarrow \mathrm{n}=8 \ & \frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{\mathrm{n}}=1 \Rightarrow \frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{8}=1 \ & \mathrm{e}=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{8}{16}}=\frac{1}{\sqrt{2}} \end{aligned}

Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P of n sides and q be the number of all quadrilaterals that can be formed by joining the vertices of P. If p + q = 126, then the eccentricity of the ellipse $\frac{x^2}{16} + \frac{y^2}{n} = 1$ is :

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