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(D) The number of triangles that can be formed is $p = {}^{n}C_{3}$.The number of quadrilaterals that can be formed is $q = {}^{n}C_{4}$.Given $p+q =

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$\frac{1}{\sqrt{2}}$

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(D) The number of triangles that can be formed is $p = {}^{n}C_{3}$.The number of quadrilaterals that can be formed is $q = {}^{n}C_{4}$.Given $p+q = 126$,we have ${}^{n}C_{3} + {}^{n}C_{4} = 126$.Using the identity ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$,we get ${}^{n+1}C_{4} = 126$.Since ${}^{9}C_{4} = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$,we have $n+1 = 9$,which implies $n = 8$.The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{8} = 1$.Here,$a^2 = 16$ and $b^2 = 8$. Since $a^2 > b^2$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.$e = \sqrt{1 - \frac{8}{16}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

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