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 x²/16 + y²/n = 1 is :

Solution

Maths sir · 4 steps

Add the two counts with Pascal's rule: C(n, 3)+C(n, 4)=C(n+1, 4), and n falls out at once.

1. Count the shapes. No three vertices of a regular polygon are collinear. So any 3 vertices give a triangle and any 4 give a quadrilateral, which makes p = C(n, 3) and q = C(n, 4).

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

2. Find n. Try values: C(9, 4) = (9 × 8 × 7 × 6)/24 = 126. So n+1 = 9 and n = 8. Check: C(8, 3)+C(8, 4) = 56 + 70 = 126.
3. Eccentricity. The ellipse is x²/16+y²/8=1, with a² = 16 along x and b² = 8. Then e² = 1 - b²/a² = 1 - 8/16 = 1/2.
4. So. e = 1/√2, which is option (a).

The idea to keep. Whenever you see C(n, r)+C(n, r+1), collapse it to C(n+1, r+1) before solving. Then just remember which value is n and which is n+1.

Answer from the NTA final answer key. Solution by PrepWiser.