Answer
a. $x'^2+9y'^2=9$
b. $\frac{x'^2}{9}+\frac{y'^2}{1}=1$
c. see graph
Work Step by Step
a. Step 1. Given
$5x^{2}-8xy+5y^{2}-9=0$
we have
$A=C=5, B=-8$
The rotation angle is
$cot(2\theta)=\frac{A-C}{B}=0, 2\theta=\frac{\pi}{2}, \theta=\frac{\pi}{4}$
Step 2. We have the transformations:
$x=\frac{\sqrt 2}{2}(x'-y')$ and $y=\frac{\sqrt 2}{2}(x'+y')$
Step 3. The equation becomes
$5x^{2}-8xy+5y^{2}-9=5(\frac{\sqrt 2}{2}(x'-y'))^{2}-8(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))+5(\frac{\sqrt 2}{2}(x'+y'))^{2}-9=x'^2+9y'^2-9=0$
or
$x'^2+9y'^2=9$
b. In standard form, we have
$\frac{x'^2}{9}+\frac{y'^2}{1}=1$
which represents an ellipse.
c. We can graph the equation as shown in the figure.
