Let m_1,m_2 be the slopes of two adjacent sides of a square of side a such that a^2+11a+3(m_1^2+m_2^2)=220. If one vertex of the square is (10( cosα…

Given {m}_{1}&{m}_{2} are slopes of two adjacent sides of square then, ${m}_{1}{m}_{2}=-1$m1​m2​=−1

Also given ${a}^{2}+11a+3({m}_{1}^{2}+{m}_{2}^{2})=220$a2+11a+3(m12​+m22​)=220

$$\Rightarrow {a}^{2}+11a+3({m}_{1}^{2}+\frac{1}{{m}_{1}^{2}})=220$$⇒a2+11a+3(m12​+m12​1​)=220

Now on plotting the diagram we get,

Now given equation of $\mathrm{AC}$AC,

$$(\mathrm{cos}\alpha -\mathrm{sin}\alpha )x+(\mathrm{sin}\alpha +\mathrm{cos}\alpha )y=10$$(cosα−sinα)x+(sinα+cosα)y=10

Now by property of square we know that diagonals are perpendicular to each other

So equation of $BD$BD will be $(\mathrm{sin}\alpha +\mathrm{cos}\alpha )x+(\mathrm{sin}\alpha -\mathrm{cos}\alpha )y+\lambda =0$(sinα+cosα)x+(sinα−cosα)y+λ=0 now is passes through $D(10(\mathrm{cos}\alpha -\mathrm{sin}\alpha ),10(\mathrm{sin}\alpha -\mathrm{cos}\alpha ))$D(10(cosα−sinα),10(sinα−cosα)),

So, equation of $BD=(\mathrm{sin}\alpha +\mathrm{cos}\alpha )x+(\mathrm{sin}\alpha -\mathrm{cos}\alpha )y=0$BD=(sinα+cosα)x+(sinα−cosα)y=0

Now slope of AB&AD will be given by angle bisector equation of AC&BD

$$\frac{(\mathrm{cos}\alpha -\mathrm{sin}\alpha )x+(\mathrm{sin}\alpha +\mathrm{cos}\alpha )y-10}{\sqrt{2}}=\pm \frac{(\mathrm{sin}\alpha +\mathrm{cos}\alpha )x+(\mathrm{sin}\alpha -\mathrm{cos}\alpha )y}{\sqrt{2}}$$2​(cosα−sinα)x+(sinα+cosα)y−10​=±2​(sinα+cosα)x+(sinα−cosα)y​

Taking positive sign we get,

$-2\mathrm{sin}\alpha x+2\mathrm{cos}\alpha y-10=0$−2sinαx+2cosαy−10=0, so slope ${m}_{1}=\mathrm{tan}\alpha$m1​=tanα

Now taking negative sign we get,

$2\mathrm{cos}\alpha x+2\mathrm{sin}\alpha y-10=0$2cosαx+2sinαy−10=0, so slope ${m}_{2}=-\text{cot}\alpha$m2​=−cotα

Now finding $a$a by using distance of point from line

Now we can see distance of $D$D from $AC$AC will be $\frac{a}{\sqrt{2}}$2​a​,

So, $\frac{10{(\mathrm{cos}\alpha -\mathrm{sin}\alpha )}^{2}+10{(\mathrm{cos}\alpha -\mathrm{sin}\alpha )}^{2}-10}{\sqrt{2}}=\frac{a}{\sqrt{2}}$2​10(cosα−sinα)2+10(cosα−sinα)2−10​=2​a​

$$\Rightarrow a=10$$⇒a=10

Now putting the value of a,{m}_{1}&{m}_{2} in ${a}^{2}+11a+3({m}_{1}^{2}+\frac{1}{{m}_{1}2})=220$a2+11a+3(m12​+m1​21​)=220 we get,

$$\Rightarrow 100+110+3({\mathrm{tan}}^{2}\alpha +{\mathrm{cot}}^{2}\alpha )=220$$⇒100+110+3(tan2α+cot2α)=220

$$\Rightarrow 210+3({\mathrm{tan}}^{2}\alpha +{\mathrm{cot}}^{2}\alpha )=220$$⇒210+3(tan2α+cot2α)=220

$$\Rightarrow 3({\mathrm{tan}}^{2}\alpha +{\mathrm{cot}}^{2}\alpha )=10$$⇒3(tan2α+cot2α)=10

$$\Rightarrow ({\mathrm{tan}}^{2}\alpha +{\mathrm{cot}}^{2}\alpha )=\frac{10}{3}$$⇒(tan2α+cot2α)=310​

So this is only possible when {\mathrm{tan}}^{2}\alpha =3&{\text{cot}}^{2}\alpha =\frac{1}{3} or vice versa

So, ${\mathrm{tan}}^{2}\alpha =3\Rightarrow \alpha =\frac{\pi }{3}$tan2α=3⇒α=3π​

So, $72({\mathrm{sin}}^{4}\alpha +{\mathrm{cos}}^{4}\alpha )+{a}^{2}-3a+13$72(sin4α+cos4α)+a2−3a+13

$$=72(\frac{9}{16}+\frac{1}{16})+100-30+13$$=72(169​+161​)+100−30+13

$$=72(\frac{5}{8})+83=45+83=128$$=72(85​)+83=45+83=128