$Theproductoftheperpediculardis\tan cedrawnfromtheorigintothepairofstraightlines6x^2-5xy-6y^2+5y-1=0is$

$\begin{array}{l}GivenEquation6x^2-5xy-6y^2+x+5y-1=0\\ Comparingwitha{x}^2+2hxy+b{y}^2+2gx+2fy+c=0\\ a=6,h=\frac{5}{2},b=-6,g=\frac{1}{2},f=\frac{5}{2},c=-1\\ \mathrm{Pr}oductofperpendiculardis\tan cefrom(0,0)\\ =\frac{\left|c\right|}{\sqrt{{(a-b)}^2+4h^2}}\\ =\frac{|-1|}{\sqrt{{(6+6)}^2+4\left(\frac{25}{4}\right)}}\\ =\frac{1}{\sqrt{144+25}}\\ =\frac{1}{\sqrt{169}}=\frac{1}{13}\end{array}$