Question

Let $m_1,m_2$ be the slopes of two adjacent sides of a square of side a such that $a^2+11a+3\left(m_2^2+m_2^2\right)=220$. 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 $72\left({\sin }^4\alpha +{\cos }^4\alpha \right)+$ $a^2-3a+13$ is equal to:

• A. 119
• B. 128
• C. 145
• D. 155

Answer 1

Answer: (B)

$m_1{m}_2=-1$
$a^2+11a+3\left(m_1^2+\frac{1}{m_1^2}\right)=220$

Eq. of $AC$
$AC=(\cos \alpha -\sin \alpha )+(\sin \alpha +\cos \alpha )y=10$
$BD=(\sin \alpha -\cos \alpha )x+(\sin \alpha -\cos \alpha )y=0$
$(10(\cos \alpha -\sin \alpha ),10(\sin \alpha -\cos \alpha ))$
$\text{ Slope of }AC=\left(\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }\right)=\tan \theta =M$
Eq. of line making an angle ${\pi }_4$ with $AC$
$m_1,m_2=\frac{m\pm \tan \frac{\pi }{4}}{1\pm m\tan \frac{\pi }{4}}$
$=\frac{m+1}{1-m}\text{ or }\frac{m-1}{1+m}$
$\frac{\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }+1}{1-\left(\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }\right)},\frac{\frac{\sin \alpha -\cos \alpha }{\sin \alpha \cos \alpha }-1}{1+\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }}$
$m_1,m_2=\tan \alpha ,\cot \alpha$
$mid\mathrm{point}\text{ of }AC\&BD$
$=M(5(\cos \alpha -\sin \alpha ),5(\cos \alpha +\sin \alpha ))$
$B(10(\cos \alpha -\sin \alpha ),10(\cos \alpha +\sin \alpha ))$
$a=AB=\sqrt{2}BM=\sqrt{2}(5\sqrt{2})=10$
$a=10$
$\because a^2+11a+3\left(m_1^2+\frac{1}{m_{1}2}\right)=220$
$100+110+3\left({\tan }^2\alpha +{\cot }^2\alpha \right)=220$
$\text{ Hence }{\tan }^2\alpha =3,{\tan }^2\alpha =\frac{1}{3}\Rightarrow \alpha =\frac{\pi }{3}\text{ or }\frac{\pi }{6}$
$\text{ Now }72\left({\sin }^4\alpha +{\cos }^4\alpha \right)+a^2-3a+13$
$=72\left(\frac{9}{16}+\frac{1}{16}\right)+100-30+13$
$=72\left(\frac{5}{8}\right)+83=45+83=128$