Answer
(a) ellipse.
(b) $\phi=\frac{\pi}{3}$. $\frac{X^2}{4}+\frac{Y^2}{9}=1$
(c) See graph.
Work Step by Step
(a) Rewrite the original equation as $21x^2+10\sqrt 3xy+31y^2+0x+0y-144=0$ and we have $A=21, B=10sqrt 3, C=31, D=E=0,F=-144$. The discriminant is $B^2-4AC=300-3604=-2304\lt0$, 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 $cot2\phi=\frac{A-C}{B}=\frac{21-31}{10\sqrt 3}=-\frac{\sqrt 3}{3}$ which gives $2\phi=\frac{2\pi}{3}$ and $\phi=\frac{\pi}{3}$.
The transformation formula gives:
$x=X\cdot cos\frac{\pi}{3} - Y\cdot sin\frac{\pi}{3}=\frac{1}{2}X-\frac{\sqrt 3}{2}Y$,
$y=X\cdot sin\frac{\pi}{3}+Y\cdot cos\frac{\pi}{3}=\frac{\sqrt 3}{2}X+\frac{1}{2}Y$
Use them in the original equation to get $21(\frac{1}{2}X-\frac{\sqrt 3}{2}Y)^2+10\sqrt 3(\frac{1}{2}X-\frac{\sqrt 3}{2}Y)(\frac{\sqrt 3}{2}X+\frac{1}{2}Y)+31(\frac{\sqrt 3}{2}X+\frac{1}{2}Y)^2+0x+0y-144=0$
Simplify this equation by multiplying 4 and combine like terms to get $144X^2+64Y^2=144\times4$ or
$\frac{X^2}{4}+\frac{Y^2}{9}=1$
(c) See graph.
