1. Count triangles and quadrilaterals from an $n$-gon

If a regular polygon has $n$ vertices, then:

• Number of triangles formed by choosing any $3$ vertices is $p = {n \choose 3}$
• Number of quadrilaterals formed by choosing any $4$ vertices is $q = {n \choose 4}$

Given: $p+q=126$ So, ${n \choose 3} + {n \choose 4} = 126$

2. Use a standard combinatorial identity

We know: ${n \choose 3} + {n \choose 4} = {n+1 \choose 4}$ Hence, ${n+1 \choose 4} = 126$

So, $\frac{(n+1)n(n-1)(n-2)}{24} = 126$ $(n+1)n(n-1)(n-2)=3024$

Now check values: ${9 \choose 4} = 126$ Thus, $n+1=9 \Rightarrow n=8$

3. Form the ellipse

The ellipse is $\frac{x^2}{16} + \frac{y^2}{n} = 1$ Substitute $n=8$: $\frac{x^2}{16} + \frac{y^2}{8} = 1$

Here, $a^2=16, \quad b^2=8$ with $a^2>b^2$.

4. Find eccentricity

For ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1,$ its eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}$

Therefore, $e=\sqrt{1-\frac{8}{16}} = \sqrt{1-\frac12} = \sqrt{\frac12} = \frac{1}{\sqrt{2}}$

5. Check options

• A: $\frac{1}{\sqrt{2}}$ ✅
• B: $\frac{1}{2}$
• C: $\frac{\sqrt{7}}{4}$
• D: $\frac{3}{4}$

So the correct option is A.