Let m_1, m_2 be the slopes of two adjacent sides of a square of side a such that [a^2+11 a…

Question

Let $m_{1}, m_{2}$ be the slopes of two adjacent sides of a square of side a such that $$a^{2}+11 a+3\left(m_{1}^{2}+m_{2}^{2}\right)=220$$. 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 $$(\cos \alpha-\sin \alpha) x+(\sin \alpha+\cos \alpha) y=10$$, then $$72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13$$ is equal to :

Accepted Answer

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, the point of intersection of the 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

$a^{2}+11 a+3\left(m_{1}^{2}+m_{2}^{2}ight)=220$

$m_{1}^{2}+m_{2}^{2}=\frac{220-100-110}{3}=\frac{10}{3}$

and $m_{1} m_{2}=-1$

Slopes of the sides are tan $\alpha$ and $-\cot \alpha$

$	an ^{2} \alpha=3$ or $\frac{1}{3}$

$72\left(\sin ^{4} \alpha+\cos ^{4} \alphaight)+a^{2}-3 a+13$

$=72 \cdot \frac{	an ^{4} \alpha+1}{\left(1+	an ^{2} \alphaight)^{2}}+a^{2}-3 a+13=128$

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