JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Permutations and Combinations

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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 :

(A) $\frac{3}{4}$
(B) $\frac{1}{2}$
(C) $\frac{7}{4}$
(D) $\frac{1}{2}$ ✓ Correct

Answer: (D)

Solution

The number of triangles is

(3 n)

and of quadrilaterals

(4 n)

, so

(3 n) + (4 n) = (4 n + 1) = 126

. Solving

\frac{( n + 1 ) n ( n - 1 ) ( n - 2 )}{24} = 126

gives

n + 1 = 9

, i.e.

n = 8

. The ellipse is

\frac{x ^{2}}{16} + \frac{y ^{2}}{8} = 1

with

a^{2} = 16, b^{2} = 8

, so

e = 1 - \frac{8}{16} = \frac{1}{2}

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option

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