Question

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 :

(1) \(\frac{3}{4}\)

(2) \(\frac{1}{2}\)

(3) \(\frac{\sqrt{7}}{4}\)

(4) \(\frac{1}{\sqrt{2}}\)

Answer 1

Correct option is: (4) \(\frac{1}{\sqrt{2}}\)  

Total trangles \(=\Rightarrow={ }^{h} C_{3}\)

Total auadrilaterals \(={ }^{\mathrm{h}} \mathrm{C}_{4}=\mathrm{q}\)

\({ }^{\mathrm{n}} \mathrm{C}_{3}+{ }^{\mathrm{n}} \mathrm{C}_{4}=126 \Rightarrow{ }^{\mathrm{n}+1} \mathrm{C}_{4}=126\)

\(\Rightarrow \mathrm{n}+1=9 \Rightarrow \mathrm{n}=8\)

\(\frac{x^{2}}{16}+\frac{y^{2}}{n}=1 \Rightarrow \frac{x^{2}}{16}+\frac{y^{2}}{8}=1\)

\(e=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{8}{16}}=\frac{1}{\sqrt{2}}\)