Answer
See graph and explanations.
Work Step by Step
Step 1. From the given equation
$2x^2+5xy+2y^2-\frac{9}{2}=0$
we can identify the coefficients
$A=C=2, B=5$
The angle of rotation needed is
$cot\theta=\frac{A-C}{B}=0$
Thus
$2\theta=90^\circ$ and $\theta=45^\circ$
Step 2. Set up the transformation:
$x=x'cos\theta-y'sin\theta=\frac{\sqrt 2}{2}(x'-y')$
and
$y=x'sin\theta+y'cos\theta=\frac{\sqrt 2}{2}(x'+y')$
Step 3. Transform the original equation:
$2(\frac{\sqrt 2}{2}(x'-y'))^2+5(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))+2(\frac{\sqrt 2}{2}(x'+y'))^2-\frac{9}{2}=0$
Step 4. Simplify to get:
$9x'^2-y'^2-9=0$
or
$\frac{x'^2}{1}-\frac{y'^2}{9}=1$
indicating a hyperbola in rotated coordinates.
Step 5. Graph the equation as shown in the figure.
