Let m₁, m₂ be the slopes of two adjacent sides of a square of side a such that a² + 11a + $3\left(m_1^2+m_2^2\right)=220.$ If one vertex of the square is $\left(10\left(cos\alpha -sin\alpha \right),10\left(sin\alpha +cos\alpha \right)\right),$ where $\alpha \in \left(0,\frac{\pi }{2}\right)$ and the equation of one diagonal is $\left(cos\alpha -sin\alpha \right)x+\left(sin\alpha +cos\alpha \right)y=10,$ then $72\left(si{n}^4\alpha +co{s}^4\alpha \right)+a^2-3a+13$ is equal to

Eqⁿ : y – 0 = tan45° (x – 9) Þ y = (x – 9)

Option (B) is correct

|r₁ – r₂| < c₁c₂ < r₁ + r₂

-> $\left|2-\sqrt[]{4{\lambda }^2-9}\right|<\left|2\lambda \right|<2+\sqrt[]{4{\lambda }^2-9}$

$\left|2\lambda \right|-2<\sqrt[]{4{\lambda }^2-9}$    

$4{\lambda }^2+4-8\left|\lambda \right|<4{\lambda }^2-9$

  $\lambda >\frac{13}{8},\lambda <-\frac{13}{8}$           

$\sqrt[]{4{\lambda }^2-9}>0$

$\lambda >\frac{3}{2},\lambda <-\frac{3}{2}$

$\lambda \in \left(-\infty ,-\frac{13}{8}\right)\cup \left(\frac{13}{8},\infty \right)$           

Now,

$\left|2-\sqrt[]{4{\lambda }^2-9}\right|<\left|2\lambda \right|$            

$4+4{\lambda }^2-9-4\sqrt[]{4{\lambda }^2-9}<4{\lambda }^2$  

$4\sqrt[]{4{\lambda }^2-9}>-5\Rightarrow \lambda \in R$  

$\lambda \in \left(-\infty ,-\frac{13}{8}\right)\cup \left(\frac{13}{8},\infty \right)$  

(y – 2) = m (x – 8)

⇒   x-intercept

⇒     $\left(\frac{-2}{m}+8\right)$

⇒   y-intercept

⇒   (–8m + 2)

⇒   OA + OB = $\frac{-2}{m^2}$  + 8 – 8m + 2

$f\text{'}\left(m\right)=\frac{2}{m^2}-8=0$  

-> $m^2=\frac{1}{4}$

-> $m=\frac{-1}{2}$

-> $f\left(\frac{-1}{2}\right)=18$

->Minimum = 18

Kindly consider the following figure

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$ $\frac{}{}\frac{}{}$

Obviously B (2, 2) satisfying condition (i)