## Precalculus (6th Edition) Blitzer

$\frac{(x')^2}{4}-\frac{(y')^2}{9}=1$
Step 1. Use the transformation formulas $x=x'cos\theta-y'sin\theta$ $y=x'sin\theta+y'cos\theta$ we have $x=x'cos30^\circ-y'sin30^\circ=\frac{1}{2}(\sqrt 3x'-y')$ $y=x'sin30^\circ+y'cos30^\circ=\frac{1}{2}(x'+\sqrt 3y')$ Step 2. The original equation becomes: $23x^2+26\sqrt 3xy-3y^2-144=23(\frac{1}{2}(\sqrt 3x'-y'))^2+26\sqrt 3(\frac{1}{2}(\sqrt 3x'-y'))(\frac{1}{2}(x'+\sqrt 3y'))-3(\frac{1}{2}(x'+\sqrt 3y'))^2-144=\frac{23}{4}(3(x')^2-2\sqrt 3x'y'+(y')^2)+\frac{13\sqrt 3}{2}(\sqrt 3(x')^2+2x'y'-\sqrt 3(y')^2)-\frac{3}{4}((x')^2+2\sqrt 3x'y'+3(y')^2)-144=36(x')^2-16(y')^2)-144=0$ Step 3. We can rewrite the above result as $\frac{(x')^2}{4}-\frac{(y')^2}{9}=1$, which represents a hyperbola.