## Precalculus (6th Edition) Blitzer

a. $4x'^2-y'^2=1$ b. $\frac{x'^2}{1/4}-\frac{y'^2}{1}=1$ c. see graph
a. Step 1. Given $11x^{2}+10\sqrt{3}xy+y^{2}-4=0$ we have $A=11, B=10, C=1$ The rotation angle is $cot(2\theta)=\frac{A-C}{B}=\frac{11-1}{10\sqrt 3}=\frac{\sqrt 3}{3}, 2\theta=\frac{\pi}{3}, \theta=\frac{\pi}{6}$ Step 2. We have the transformations (see also exercise 19): $x=\frac{1}{2}(\sqrt 3x'-y')$ and $y=\frac{1}{2}(x'+\sqrt 3y')$ Step 3. The equation becomes $11x^{2}+10\sqrt{3}xy+y^{2}-4=11(\frac{1}{2}(\sqrt 3x'-y'))^{2}+10\sqrt{3}(\frac{1}{2}(\sqrt 3x'-y'))(\frac{1}{2}(x'+\sqrt 3y'))+(\frac{1}{2}(x'+\sqrt 3y'))^{2}-4=16x'^2-4y'^2-4=0$ or $4x'^2-y'^2=1$ b. In standard form, we have $\frac{x'^2}{1/4}-\frac{y'^2}{1}=1$ which represents a hyperbola. c. We can graph the equation as shown in the figure.