Question

Let m1, m2 be the slopes of two adjacent sides of a square of side a such that $a^2 + 11a + 3(m_1^2 + m_2^2) = 220$. 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 $(\cos \alpha - \sin \alpha)x + (\sin \alpha + \cos \alpha)y = 10$, then $72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13$ is equal to :

• A. 119
• B. 128
• C. 145
• D. 155

Answer 1

Answer: (B)

128

Answer 2

One vertex of square is
$(10(\cos \alpha - \sin \alpha), 10(\sin \alpha + \cos \alpha))$
and one of the diagonal is
$(\cos \alpha - \sin \alpha)x + (\sin \alpha + \cos \alpha)y = 10$

So the other diagonal can be obtained as
$(\cos \alpha + \sin \alpha)x - (\cos \alpha - \sin \alpha)y = 0$

So, point of intersection of diagonal will be
$(5(\cos \alpha - \sin \alpha), 5(\cos \alpha + \sin \alpha))$

Therefore, the vertex opposite to the given vertex is (0, 0).
So, the diagonal length
$10\sqrt{2}$
Side length $(a) = 10$

It is given that
$a^2 + 11a + 3(m_1^2 + m_2^2) = 220$
$m_1^2 + m_2^2 = 220 - 100 - 110 = 103$
and $m_1 \cdot m_2 = -1$

Slopes of the sides are $\tan \alpha \quad \text{and} \quad -\cot \alpha$
$\tan(2\alpha) = 3 \quad \text{or} \quad \tan(2\alpha) = \frac{1}{3}$

$72(\sin 4\alpha + \cos 4\alpha) + a^2 - 3a + 13$
$72 \cdot \tan^4 \alpha + \frac{1}{{(1 + \tan^2 \alpha)}^2} + a^2 - 3a + 13 = 128$
So, the correct option is (B): $128$