Answer
$\frac{(x')^2}{9}+\frac{(y')^2}{4}=1$
Work Step by Step
Step 1. Using the transformation formulas
$x=x'cos\theta-y'sin\theta$
$y=x'sin\theta+y'cos\theta$
we have
$x=x'cos45^\circ-y'sin45^\circ=\frac{\sqrt 2}{2}(x'-y')$
$y=x'sin45^\circ+y'cos45^\circ=\frac{\sqrt 2}{2}(x'+y')$
Step 2. The original equation becomes:
$13x^2-10xy+13y^2-72=13(\frac{\sqrt 2}{2}(x'-y'))^2-10(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))+13(\frac{\sqrt 2}{2}(x'+y'))^2-72=\frac{13}{2}((x')^2-2x'y'+(y')^2)-5((x')^2-(y')^2)+\frac{13}{2}((x')^2+2x'y'+(y')^2)-72=8(x')^2+18(y')^2-72=0$
Step 3. We can rewrite the above result as $\frac{(x')^2}{9}+\frac{(y')^2}{4}=1$, which represents an ellipse.
