Let m₁,m₂ be the slopes of two adjacent sides of a square of side a such that a²+11a+3(m₁²+m₂²)=220. If one vertex of the square is (10(cos α-sin α),10(sin α+cos α)), where α∈ (0,π/2) and the equation of one diagonal is (cos α-sin α)x+(sin α+cos α)y=10, then 72(sin⁴ α+cos⁴ α)+a²-3a+13 is equal to :

Solution

Maths sir · 4 steps

The vertex is off the given diagonal, so its distance to it is half the diagonal, and that fixes a.

1. Find the side. Put the vertex into the diagonal's left side: 10(cos α-sin α)²+10(sin α+cos α)² = 20 ≠ 10. So this vertex is not on the given diagonal. Its distance to that diagonal is half the other diagonal, a√2/2 = a/√2. The coefficients of the line give √((cos α-sin α)²+(sin α+cos α)²) = √2.

   a/√2 = (|20-10|)/√2 ⇒ a = 10

2. Find the slopes. With a = 10: 100 + 110 + 3(m₁²+m₂²) = 220, so m₁²+m₂² = 10/3. Adjacent sides are perpendicular, so m₁ m₂ = -1. That gives m₁² + 1/m₁² = 10/3, so the slopes are ±√3 and ∓1/√3.
3. Link slopes to α. The diagonal's normal is (cos α-sin α, sin α+cos α) = √2(cos(α+45°), sin(α+45°)), so the diagonal points along α+135°. The sides make 45° with it, so they point along α and α+90°, with slopes tan α and -cot α. So tan α = √3 or 1/√3, and in both cases sin 2α = √3/2.
4. So. sin⁴ α+cos⁴ α = 1 - 1/2 sin² 2α = 1 - 3/8 = 5/8, and 72 × 5/8 = 45. Also a² - 3a + 13 = 100 - 30 + 13 = 83. The total is 45 + 83 = 128, option (b).

The idea to keep. In a square, a vertex not on a diagonal sits exactly half a diagonal, a/√2, away from it. And the sides always make 45° with the diagonals, which links their slopes to the diagonal's angle.

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