Question

Identify the graph of the equation. What is the angle of rotation for the equation?

2xy – 9 = 0

Answer 1

b. hyperbola,
`45\degree`

Explanation:

The given equation is;

`2xy-9=0`

Comparing to the general equation;

`Ax^2+Bxy+Cy^2+Dx+Ey+F=0`,

`A=0,B=2,C=0`

We can eliminate the xy term using;

`\cot(2\theta)=(A-C)/(B)`

`\cot(2\theta)=(0-0)/(2)`

`\Rightarrow \cot(2\theta)=0`

`\Rightarrow 2\theta=\cot^(-1)(0)`

`\Rightarrow 2\theta=90\degree`

`\Rightarrow \theta=45\degree`

This implies that;

`\cos(\theta)=\sin(\theta)=(1)/(√(2) )`

`x=(x')/(√(2) ) -(y')/(√(2) )`

`y=(x')/(√(2) ) +(y')/(√(2) )`

Substitute into the original equation;

`2((x')/(√(2) ) -(y')/(√(2) ))((x')/(√(2) ) +(y')/(√(2) ))-9=0`

`2(((x')^2)/(2) -((y')^2)/(2))-9=0`

`(x')^2 -(y')^2=9`

`((x')^2)/(9) -((y')^2)/(9))=1`

This is a hyperbola;