Assertion (A) : The slopes of one line represented by $2x^2-5xy+2y^2=0$ is 4 times the slope of the second line.

               Reason( R): If the slopes of lines represented by $a{x}^2+2hxy+b{y}^2=0$ are in m : n

then $\frac{{\left(m+n\right)}^2}{mn}=\frac{4h^2}{ab}$

Assertion (A ) : Given Equation $2x^2-5xy+2y^2=0$

Comparing with

$\begin{array}{l}a{x}^2+2hxy+b{y}^2=0\\ a=2,h=\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,b=2\\ Lettheratioofslopem:n=4:1\\ \frac{{\left(m+n\right)}^2}{mn}=\frac{{\left(4+1\right)}^2}{4\left(1\right)}=\frac{25}{4}\\ and\frac{4h^2}{ab}=\frac{4\left(\frac{25}{4}\right)}{2\left(2\right)}=\frac{25}{4}\\ \therefore \frac{{\left(m+n\right)}^2}{mn}=\frac{4h^2}{ab}\\ True.\end{array}$

$\begin{array}{l}\mathrm{Re}ason\left(R\right):Let{m}_1=mkand{m}_2=nk\\ Then{m}_1+m_2=\frac{-2h}{b},m_1{m}_2=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.\\ \Rightarrow \left(m+n\right)k=\frac{-2h}{b},\Rightarrow mn{k}^2=\frac{a}{b}\\ \Rightarrow {\left(m+n\right)}^2{k}^2=\frac{4h^2}{b^2},\therefore k^2=\frac{a}{bmn}\\ \therefore {\left(m+n\right)}^2.\left(\frac{a}{bmn}\right)=\frac{4h^2}{b^2}\\ \Rightarrow \frac{{\left(m+n\right)}^2}{mn}=\frac{4h^2}{b^2}\Rightarrow Ristrue.\end{array}$