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JEE Mathematics 2025 Question with Solution

Let pp be the number of all triangles that can be formed by joining the vertices of a regular polygon PP of nn sides, and qq be the number of all quadrilaterals that can be formed by joining the vertices of PP. If p+q=126p + q = 126, then the eccentricity of the ellipse x216+y2n=1\frac{x^2}{16} + \frac{y^2}{n} = 1 is:

  • A

    34\frac{3}{4}

  • B

    12\frac{1}{2}

  • C

    74\frac{\sqrt{7}}{4}

  • D

    12\frac{1}{\sqrt{2}}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: pp is the number of triangles and qq is the number of quadrilaterals formed from the vertices of a regular polygon of nn sides. Also, p+q=126p+q=126 and the ellipse is

x216+y2n=1\frac{x^2}{16}+\frac{y^2}{n}=1

Find: The eccentricity of the ellipse.

The number of triangles is

p=(n3)=n(n−1)(n−2)6p=\binom{n}{3}=\frac{n(n-1)(n-2)}{6}

and the number of quadrilaterals is

q=(n4)=n(n−1)(n−2)(n−3)24q=\binom{n}{4}=\frac{n(n-1)(n-2)(n-3)}{24}

Using p+q=126p+q=126,

n(n−1)(n−2)6+n(n−1)(n−2)(n−3)24=126\frac{n(n-1)(n-2)}{6}+\frac{n(n-1)(n-2)(n-3)}{24}=126

Multiplying by 2424,

4n(n−1)(n−2)+n(n−1)(n−2)(n−3)=30244n(n-1)(n-2)+n(n-1)(n-2)(n-3)=3024

Factorising,

n(n−1)(n−2)(4+n−3)=3024n(n-1)(n-2)\big(4+n-3\big)=3024 n(n−1)(n−2)(n+1)=3024n(n-1)(n-2)(n+1)=3024

Testing integer values, for n=8n=8,

8⋅7⋅6⋅9=30248\cdot 7\cdot 6\cdot 9=3024

So, n=8n=8.

Substituting in the ellipse,

x216+y28=1\frac{x^2}{16}+\frac{y^2}{8}=1

Hence,

a2=16,b2=8a^2=16, \quad b^2=8

The eccentricity is

e=1−b2a2e=\sqrt{1-\frac{b^2}{a^2}}

Therefore,

e=1−816=12=12e=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}

Therefore, the eccentricity of the ellipse is 12\frac{1}{\sqrt{2}} and the correct option is D.

Using combinations explicitly

Given: A regular polygon has nn vertices. Any triangle is formed by choosing any 33 vertices, and any quadrilateral is formed by choosing any 44 vertices. Find: The eccentricity after determining nn from p+q=126p+q=126.

Since no three chosen vertices of a regular polygon are collinear, every choice of 33 vertices gives a triangle. Thus,

p=(n3)p=\binom{n}{3}

Similarly, every choice of 44 vertices gives a quadrilateral, so

q=(n4)q=\binom{n}{4}

Now,

(n3)+(n4)=126\binom{n}{3}+\binom{n}{4}=126

Substituting the formulas,

n(n−1)(n−2)6+n(n−1)(n−2)(n−3)24=126\frac{n(n-1)(n-2)}{6}+\frac{n(n-1)(n-2)(n-3)}{24}=126

Taking n(n−1)(n−2)24\frac{n(n-1)(n-2)}{24} common,

n(n−1)(n−2)24(4+n−3)=126\frac{n(n-1)(n-2)}{24}\left(4+n-3\right)=126 n(n−1)(n−2)(n+1)24=126\frac{n(n-1)(n-2)(n+1)}{24}=126 n(n−1)(n−2)(n+1)=3024n(n-1)(n-2)(n+1)=3024

Checking consecutive values,

8⋅7⋅6⋅9=30248\cdot 7\cdot 6\cdot 9=3024

Hence,

n=8n=8

So the ellipse becomes

x216+y28=1\frac{x^2}{16}+\frac{y^2}{8}=1

Comparing with the standard form

x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

we get

a2=16,b2=8a^2=16, \quad b^2=8

Then

e=1−b2a2=1−816=12=12e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}

Thus, the correct option is D.

Common mistakes

  • Using q=(n2)q=\binom{n}{2} or another incorrect combination count for quadrilaterals is wrong because a quadrilateral needs 4 vertices, not 2. Count quadrilaterals with q=(n4)q=\binom{n}{4}.

  • Stopping at the incorrect trial value n=9n=9 is wrong because it does not satisfy

    n(n−1)(n−2)(n+1)=3024n(n-1)(n-2)(n+1)=3024

    Verify the value by substitution; the correct value is obtained at n=8n=8.

  • Applying the eccentricity formula with the larger denominator in the wrong place is incorrect. For an ellipse in standard form, use a2>b2a^2>b^2 and then compute e=1−b2a2e=\sqrt{1-\frac{b^2}{a^2}}.

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