## Precalculus (6th Edition) Blitzer

$(0,\pm\sqrt 6)$
Step 1. Given $2x^{2}-4xy+5y^{2}-36=0$, we have $A=2, B=-4, C=5$ and the rotation angle is $cot(2\theta)=\frac{A-C}{B}=\frac{2-5}{-4}=\frac{3}{4}$ Step 2. We have the transformations (see also exercise 15): $x=\frac{\sqrt 5}{5}(2x'-y')$ and $y=\frac{\sqrt 5}{5}(x'+2y')$ Step 3. The equation becomes $2x^{2}-4xy+5y^{2}-36=2(\frac{\sqrt 5}{5}(2x'-y'))^{2}-4(\frac{\sqrt 5}{5}(2x'-y'))(\frac{\sqrt 5}{5}(x'+2y'))+5(\frac{\sqrt 5}{5}(x'+2y'))^{2}-36=x'^2+6y'^2-36=0$ or, $x'^2+6y'^2=36$ Step 4. In standard form, we have $\frac{x'^2}{36}+\frac{y'^2}{6}=1$, which represents an ellipse. Step 5. We can find the endpoints of the minor axis in the $x'-y'$ system as $(0,\pm\sqrt 6)$