Let \(\mathrm{m}_{1}, \mathrm{m}_{2}\) be the slopes of two adjacent sides of a square of side a such that \(\mathrm{a}^{2}+11\mathrm{a}+3\left(\mathrm{m}_{1}^{2}+\mathrm{m}_{2}^{2}\right)=220\). If one vertex of the square is \((10(\cos\alpha-\sin\alpha), 10(\sin\alpha+\cos\alpha))\), where \(\alpha \in\left(0, \frac{\pi}{2}\right)\) and the equation of one diagonal is \((\cos\alpha-\sin\alpha)\mathrm{x}+(\sin\alpha+\cos\alpha)\mathrm{y}=10\), then \(72\left(\sin^{4}\alpha+\cos^{4}\alpha\right)+a^{2}-3a+13\) is equal to:

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Straight Lines

Let

m_{1}, m_{2}

be the slopes of two adjacent sides of a square of side a such that

a^{2} + 11 a + 3 (m_{1}^{2} + m_{2}^{2}) = 220

. If one vertex of the square is

(10 (cos α - sin α), 10 (sin α + cos α))

, where

α \in (0, \frac{π}{2})

and the equation of one diagonal is

(cos α - sin α) x + (sin α + cos α) y = 10

, then

72 (sin^{4} α + cos^{4} α) + a^{2} - 3 a + 13

is equal to:

Answer: (B)

Since adjacent sides of a square are perpendicular,

m_{1} m_{2} = - 1

. The slopes of sides making

4 5^{\circ}

with the diagonal are

m_{1} = tan α

and

m_{2} = cot α

. From the geometry, the side length

a = 10

. Substituting into

a^{2} + 11 a + 3 (m_{1}^{2} + m_{2}^{2}) = 220

100 + 110 + 3 (tan^{2} α + cot^{2} α) = 220

, giving

tan^{2} α + cot^{2} α = 10/3

. This yields

tan^{2} α = 3

1/3

, so

α = π /3

π /6

. For either value,

sin^{4} α + cos^{4} α = 5/8

. Thus

72 (5/8) + 100 - 30 + 13 = 45 + 83 = 128

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option