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

The Correct Option is B

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): \(128\)