Answer
(a) Ellipse.
(b) $\phi=\frac{\pi}{6}$, $(X-1)^2+4Y^2=1$
(c) See graph.
Work Step by Step
(a) Write the equation as $7x^2-6\sqrt 3xy+13y^2-4\sqrt 3x-4y=0$, we can identify $A=7, B=-6\sqrt 3, C=13$. The discriminant is $B^2-4AC=3108-364=-256$, thus the graph of the equation will be an ellipse.
(b) Step 1. To eliminate the xy-term, we need to rotate the axes with an angle $\phi$ given by $cot2\phi=\frac{A-C}{B}=\frac{7-13}{-6\sqrt 3}=\frac{\sqrt 3}{3}$ which gives $2\phi=\frac{\pi}{3}$ and $\phi=\frac{\pi}{6}$
Step 2. Use the conversion formulas $x=Xcos\phi-Ysin\phi=\frac{\sqrt 3}{2}X-\frac{1}{2}Y$ and $y=Xsin\phi+Ycos\phi=\frac{1}{2}X+\frac{\sqrt 3}{2}Y$
Step 3. Plug the above relation into the original equation to get $7(\frac{\sqrt 3}{2}X-\frac{1}{2}Y)^2-6\sqrt 3(\frac{\sqrt 3}{2}X-\frac{1}{2}Y)(\frac{1}{2}X+\frac{\sqrt 3}{2}Y)+13(\frac{1}{2}X+\frac{\sqrt 3}{2}Y)^2-4\sqrt 3(\frac{\sqrt 3}{2}X-\frac{1}{2}Y)-4(\frac{1}{2}X+\frac{\sqrt 3}{2}Y)=0$
Step 4. Multiply 4 to both sides and simplify the equation to get $16X^2+64Y^2-32X=0$ or $X^2-2X+1+4Y^2=1$ which gives $(X-1)^2+4Y^2=1$
(c) See graph.
