1.Directrices of 3x2−3y2−18x+12y+2=03x^2-3y^2-18x+12y+2=03x2−3y2−18x+12y+2=0 area.x=3±13/6x = 3 \pm \sqrt{13/6}x=3±13/6b.x=3±6/13x = 3 \pm \sqrt{6/13}x=3±6/13c.x=6±13/3x = 6 \pm \sqrt{13/3}x=6±13/3d.x=6±3/13x = 6 \pm \sqrt{3/13}x=6±3/13Login to continueOnly logged in users canattempt or see the solution.
