અહી m_{1}, m_{2} એ ચોરસની પાસપાસને બાજુઓના ઢા | vedclass

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Question

$m_{1} m_{2}=-1$

$a^{2}+11 a+3\left(m_{1}^{2}+\frac{1}{m_{1}^{2}}\right)=220$

Eq. of $AC$

$AC =(\cos \alpha-\sin \alpha)+(\sin \alpha+\cos \a

Vedclass · Accepted Answer

$128$

Vedclass · Answer

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

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