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;
