Question

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

• A. 119
• B. 128
• C. 145
• D. 155

Answer 1

Answer: (B)

$ m_1 m_2=-1 $
$ a^2+11 a+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 \operatorname{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+11 a +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 ^2-3 a +13$
$=72\left(\frac{9}{16}+\frac{1}{16}\right)+100-30+13 $
$=72\left(\frac{5}{8}\right)+83=45+83=128$