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