Answer
(a) hyperbola.
(b) $\phi=\frac{\pi}{6}$.
$3X^2-Y^2=2\sqrt 3$
(c) See graph.
Work Step by Step
(a) Rewrite the original equation as $\sqrt 3x^2+3xy+0y^2+0x+0y-3=0$ and we have $A=\sqrt 3, B=3, C=0, D=E=0,F=-3$. The discriminant is $B^2-4AC=9\gt0$, thus the graph of the equation is a hyperbola.
(b) To eliminate the xy-term, we need a rotation of axes with an angle $\phi$ where $cot2\phi=\frac{A-C}{B}=\frac{\sqrt 3}{3}$ which gives $2\phi=\frac{\pi}{3}$ and $\phi=\frac{\pi}{6}$. The transformation formula gives:
$x=X\cdot cos\frac{\pi}{6} - Y\cdot sin\frac{\pi}{6}=\frac{\sqrt 3}{2}X-\frac{1}{2}Y$,
$y=X\cdot sin\frac{\pi}{6}+Y\cdot cos\frac{\pi}{6}=\frac{1}{2}X+\frac{\sqrt 3}{2}Y$
Use them in the original equation to get $\sqrt 3(\frac{\sqrt 3}{2}X-\frac{1}{2}Y)^2+3(\frac{\sqrt 3}{2}X-\frac{1}{2}Y)(\frac{1}{2}X+\frac{\sqrt 3}{2}Y)-3=0$
Simplify this equation by multiplying 4 and combine the like terms to get $3\sqrt 3X^2-\sqrt 3Y^2=6$ or
$3X^2-Y^2=2\sqrt 3$
(c) See graph.
