Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P on 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 ellip

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: [2025]

Smart Achievers · Accepted Answer

(3)

Total triangles $= {}^{n}C_{3}$

Total quadrilaterals $= {}^{n}C_{4}$

$∵ {}^{n}C_{3} + {}^{n}C_{4} = 126 \Rightarrow {}^{n + 1}C_{4} = 126$

$\Rightarrow n = 8$

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

$∴ e = \sqrt{1 - \frac{8}{16}} = \sqrt{\frac{8}{16}} = \frac{1}{\sqrt{2}}$

Solution

1 $\frac{\sqrt{7}}{4}$

2 $\frac{1}{2}$

3 $\frac{1}{\sqrt{2}}$

4 $\frac{3}{4}$

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Solution

Q. 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 x216+y2n=1 is: [2025]

1 74

2 12

3 12

4 34

Ans. (3) Total triangles =C3n Total quadrilaterals =C4n ∵ C3n+C4n=126 ⇒ C4n+1=126 ⇒ n=8 The ellipse is x216+y2n=1 ⇒ x216+y28=1 ∴ e=1–816=816=12

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1 $\frac{\sqrt{7}}{4}$

2 $\frac{1}{2}$

3 $\frac{1}{\sqrt{2}}$

4 $\frac{3}{4}$