Identify the graph of the equation. What is the angle of rotation for the equation?2xy – 9 = 0

Answer:b. hyperbola, [tex]45\degree[/tex]Step-by-step explanation:The given equation is;[tex]2xy-9=0[/tex]Comparing to the general equation;[tex]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/tex],[tex]A=0,B=2,C=0[/tex]We can eliminate the xy term using;[tex]\cot(2\theta)=\frac{A-C}{B}[/tex][tex]\cot(2\theta)=\frac{0-0}{2}[/tex][tex]\Rightarrow \cot(2\theta)=0[/tex][tex]\Rightarrow 2\theta=\cot^{-1}(0)[/tex][tex]\Rightarrow 2\theta=90\degree[/tex][tex]\Rightarrow \theta=45\degree[/tex]This implies that;[tex]\cos(\theta)=\sin(\theta)=\frac{1}{\sqrt{2} }[/tex][tex]x=\frac{x'}{\sqrt{2} } -\frac{y'}{\sqrt{2} }[/tex][tex]y=\frac{x'}{\sqrt{2} } +\frac{y'}{\sqrt{2} }[/tex]Substitute into the original equation;[tex]2(\frac{x'}{\sqrt{2} } -\frac{y'}{\sqrt{2} })(\frac{x'}{\sqrt{2} } +\frac{y'}{\sqrt{2} })-9=0[/tex][tex]2(\frac{(x')^2}{2} -\frac{(y')^2}{2})-9=0[/tex][tex](x')^2 -(y')^2=9[/tex][tex]\frac{(x')^2}{9} -\frac{(y')^2}{9})=1[/tex]This is a hyperbola;