- Home
- Standard 11
- Mathematics
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.
$119$
$128$
$145$
$155$
Solution

$m_{1} m_{2}=-1$
$a^{2}+11 a+3\left(m_{1}^{2}+\frac{1}{m_{1}^{2}}\right)=220$
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 $AC$
$m _{1}, m _{2}=\frac{ m \pm tan }{1 \pm m \operatorname{ta}}$
$=\frac{ m +1}{1- m }$ 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 point of $AC \& BD$
$= M (5(\cos \alpha-\sin \alpha), 5(\cos \alpha+\sin \alpha))$
$B (10(\cos \alpha-\sin \alpha), 10(\cos \alpha+\sin \alpha))$
$a = AB =\sqrt{2} BM =\sqrt{2}(5 \sqrt{2})=10$
$a =10$
$\because a^{2}+11 a+3\left(m_{1}^{2}+\frac{1}{m_{1} 2}\right)=220$
$100+110+3\left(\tan ^{2} \alpha+\cot ^{2} \alpha\right)=220$
Hence $\tan ^{2} \alpha=3, \tan ^{2} \alpha=\frac{1}{3} \Rightarrow \alpha=\frac{\pi}{3}$ or $\frac{\pi}{6}$
Now $72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13$
$=72\left(\frac{9}{16}+\frac{1}{16}\right)+100-30+13$
$=72\left(\frac{5}{8}\right)+83=45+83=128$