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Let \mathrm{ m_{1}, m_{2}} be the slopes of two adjacent sides of a square of side a such that \mathrm{a^{2}+11 a+3\left(m_{1}^{2}+m_{2}^{2}\right)=220 }.
If one vertex of the square is\mathrm{(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))}, where \mathrm{\alpha \in\left(0, \frac{\pi}{2}\right)} and the equation of one diagonal is \mathrm{(\cos \alpha-\sin \alpha) x+(\sin \alpha+\cos \alpha) y=10}, then\mathrm{72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13} is equal to :

Option: 1

119


Option: 2

128


Option: 3

145


Option: 4

155


Answers (1)

best_answer

\begin{aligned} &\mathrm{\frac{a}{\sqrt{2}}=\frac{\left | 10\left ( \cos \alpha -\sin \alpha \right )^{2}+10\left ( \cos \alpha +\sin \alpha \right )^{2}-10 \right |}{\sqrt{\left ( \cos \alpha -\sin \alpha \right )^{2}+\left ( \cos \alpha +\sin \alpha \right )^{2}}}}\\ &\Rightarrow \frac{a}{\sqrt{2}}=\frac{10}{\sqrt{2}}\\ &\Rightarrow a=10\\ &\therefore \text{Equation } \mathrm{a^{2}+11a+(m_{1}^{2}+m_{2}^{2})=220}\\ &\Rightarrow \mathrm{100+110+3(m_{1}^{2}+m_{2}^{2})=220}\\ &\mathrm{3\left ( m_{1}^{2}+m_{2}^{2} \right )=10} \end{aligned}

\begin{aligned} &\mathrm{\Rightarrow m_{1}^{2}+m_{2}^{2}=\frac{10}{3}} \\ &\mathrm{\Rightarrow m_{1}^{2}+\frac{1}{m_{1}^{2}}=\frac{10}{3}} \quad \text { (As }\mathrm{ m_{1} m_{2}=-1} \text { ) } \\ & \mathrm{\Rightarrow m_{1}=\frac{1}{\sqrt{3}} \text { or }-\frac{1}{\sqrt{3}}} \text { or }\mathrm{ \sqrt{3}} \text { or }\mathrm{-\sqrt{3}} \end{aligned}

\begin{aligned} &\mathrm{ \therefore\left(m_{1}, m_{2}\right) is \left(\frac{1}{\sqrt{3}},-\sqrt{3}\right) or \left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right)}\\ &\text{For}\mathrm{\left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right)},\text{ sides make}\mathrm{ -30^{\circ} }\text{ and } \mathrm{60^{\circ}} \text{ with } \mathrm{+x} \text{ axis , therefore diagonal makes } \mathrm{15^{\circ} } \text{ with }\mathrm{ +x } \text{ axis }\\ &\text{ Also given slope of diagonal } \mathrm{=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}} \end{aligned}

\begin{aligned} &\mathrm{\Rightarrow \frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}=\tan 15^{\circ} }\\ &\mathrm{\Rightarrow \quad \frac{\tan \alpha-1}{1+\tan \alpha}=\tan 15^{\circ}} \\ &\mathrm{\Rightarrow \quad \tan \left(\alpha-45^{\circ}\right)=\tan _{1} 15^{\circ} }\\ &\mathrm{\Rightarrow \quad \alpha=60^{\circ} }\\ &\mathrm{\therefore \quad 72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13}\\ &\mathrm{=72\left(\frac{9}{16}+\frac{1}{16}\right)+100-30+13} \\ &\mathrm{=128} \\ &\therefore \text { Option (B)} \end{aligned}

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