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 :
- 119
- 128
- 145
- 155
The Correct Option is B
Approach Solution - 1
To solve the problem, we need to determine the expression \(72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13\), given a set of conditions related to a square.
- From the problem, we have the equation: \(a^2 + 11a + 3(m_1^2 + m_2^2) = 220\).
- We are given that \((\cos \alpha - \sin \alpha) x + (\sin \alpha + \cos \alpha) y = 10\) is the equation of the diagonal of the square. The slope of this line is: \(-\frac{(\cos \alpha - \sin \alpha)}{(\sin \alpha + \cos \alpha)}\).
- The vertex of the square is \((10(\cos \alpha - \sin \alpha), 10(\sin \alpha + \cos \alpha))\). Simplifying using trigonometric identities:
- We use identities: \(\sin^2 \alpha + \cos^2 \alpha = 1\), \((\sin \alpha + \cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos \alpha\) to deduce: \((\sin^2 \alpha + \cos^2 \alpha) + 2\sin \alpha \cos \alpha = 1 + 2\sin \alpha \cos \alpha = (\sin \alpha + \cos \alpha)^2\).
- The condition \(a^2 + 11a + 3(m_1^2 + m_2^2) = 220\) means the diagonals of the square intersect at the point given.
- Using the equation of the diagonal: \((\cos \alpha - \sin \alpha) (10(\cos \alpha - \sin \alpha)) + (\sin \alpha + \cos \alpha) (10(\sin \alpha + \cos \alpha)) = 10\), solving it yields: \(10((\cos \alpha - \sin \alpha)^2 + (\sin \alpha + \cos \alpha)^2) = 10\), further simplifies to: \(10(1 + 2\sin \alpha \cos \alpha) = 10\).
- From the relation: \(a^2 + 11a + 220 - a^2 - 11a = 0 \Rightarrow m_1^2 + m_2^2 = \frac{200 - a^2 - 11a}{3}\).
- To find: \(72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13\), use the identity: \((\sin^2 \alpha + \cos^2 \alpha)^2 = \sin^4 \alpha + \cos^4 \alpha + 2\sin^2 \alpha \cos^2 \alpha = 1\) to find: \(72(\sin^4 \alpha + \cos^4 \alpha) = 18(1 - 2\sin^2 \alpha \cos^2 \alpha) = 72 - 36 \sum\).
- Finally, simplify: \(72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13 = 128\).
Therefore, the answer is 128.
Approach Solution -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): \(128\)
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