Answer
a. $x'^2+6y'=0$
b. $x'^2=-6y'$ .
c. see graph
Work Step by Step
a. Step 1. Given
$x^{2}+2\sqrt 3xy+3y^{2}-12\sqrt 3x+12y=0$
we have
$A=1, B=2\sqrt 3, C=3$
and the rotation angle is
$cot(2\theta)=\frac{A-C}{B}=\frac{1-3}{2\sqrt 3}=-\frac{\sqrt 3}{3}, 2\theta=\frac{2\pi}{3}, \theta=\frac{\pi}{3}$,
Step 2. We have the transformations (see also section 9.4 exercise 34):
$x=\frac{1}{2}(x'-\sqrt 3y')$ and $y=\frac{1}{2}(\sqrt 3x'+y')$
Step 3. The equation becomes
$x^{2}+2\sqrt 3xy+3y^{2}-12\sqrt 3x+12y=(\frac{1}{2}(x'-\sqrt 3y'))^{2}+2\sqrt 3(\frac{1}{2}(x'-\sqrt 3y'))(\frac{1}{2}(\sqrt 3x'+y'))+3(\frac{1}{2}(\sqrt 3x'+y'))^{2}-12\sqrt 3(\frac{1}{2}(x'-\sqrt 3y'))+12(\frac{1}{2}(\sqrt 3x'+y'))=4x'^2+24y'=0$
or, $x'^2+6y'=0$
b. In standard form, we have $x'^2=-6y'$, which represents a parabola.
c. We can graph the equation as shown in the figure
