Answer
(a) Hyperbola.
(b) $\phi=\frac{\pi}{4}$, $3x^2-y^2=1$
(c) See graph.
Work Step by Step
(a) Rewrite the original equation as $x^2+4xy+y^2+0x+0y-1=0$ and we get $A=1, B=4, C=1$. The discriminant is $B^2-4AC=16-4=12\gt0$, thus the graph of the equation is a hyperbola.
(b) Step 1. To eliminate the xy-term, we need to rotate the axes with an angle $\phi$ given by $cot2\phi=\frac{A-C}{B}=0$ and we get $2\phi=\frac{\pi}{2}$ or $\phi=\frac{\pi}{4}$
Step 2. Recall the conversion formulas $x=Xcos\phi-Ysin\phi=\frac{\sqrt 2}{2}(X-Y)$ and $y=Xsin\phi+Ycos\phi=\frac{\sqrt 2}{2}(X+Y)$
Step 3. Plug the above relations into the original equation to get
$(\frac{\sqrt 2}{2}(X-Y))^2+4(\frac{\sqrt 2}{2}(X-Y))(\frac{\sqrt 2}{2}(X+Y))+(\frac{\sqrt 2}{2}(X+Y))^2-1=0$
Step 4. Multiply 2 on both sides and simplify the above equation to get
$6x^2-2y^2-2=0$ or $3x^2-y^2-1=0$
(c) See graph.
