Answer
(a) Ellipse. (b) $\phi=\frac{\pi}{6}$. $X^2+\frac{Y^2}{4}=1$ (c) See graph.
Work Step by Step
(a) Rewrite the original equation as $13x^2+6\sqrt 3xy+7y^2+0x+0y-16=0$ and we have $A=13, B=6\sqrt 3, C=7, D=E=0,F=-16$. The discriminant is $B^2-4AC=108-364=-256\lt0$, thus the graph of the equation is an ellipse.
(b) To eliminate the xy-term, we need a rotation of axes with an angle $\phi$ where $cot2\phi=\frac{A-C}{B}=\frac{13-7}{6\sqrt 3}=\frac{\sqrt 3}{3}$ 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 $13(\frac{\sqrt 3}{2}X-\frac{1}{2}Y)^2+6\sqrt 3(\frac{\sqrt 3}{2}X-\frac{1}{2}Y)(\frac{1}{2}X+\frac{\sqrt 3}{2}Y)+7(\frac{1}{2}X+\frac{\sqrt 3}{2}Y)^2-16=0$
Simplify the equation to get $4X^2+Y^2=4$ or $X^2+\frac{Y^2}{4}=1$ ]
(c) See graph.
