Answer
a. $x'^2+4y'^2=4$
b. $\frac{x'^2}{4}+\frac{y'^2}{1}=1$
c. see graph
Work Step by Step
a. Step 1. Given
$7x^{2}-6\sqrt{3}xy+13y^{2}-16=0$
we have
$A=7, B=-6\sqrt 3, C=13$
and the rotation angle is
$cot(2\theta)=\frac{A-C}{B}=\frac{7-13}{-6\sqrt 3}=\frac{\sqrt 3}{3}, 2\theta=\frac{\pi}{3}, \theta=\frac{\pi}{6}$
Step 2. We have the transformations (see also exercise 12):
$x=\frac{1}{2}(\sqrt 3x'-y')$ and $y=\frac{1}{2}(x'+\sqrt 3y')$
Step 3. The equation becomes
$7x^{2}-6\sqrt{3}xy+13y^{2}-16=7(\frac{1}{2}(\sqrt 3x'-y'))^{2}-6\sqrt{3}(\frac{1}{2}(\sqrt 3x'-y'))(\frac{1}{2}(x'+\sqrt 3y'))+13(\frac{1}{2}(x'+\sqrt 3y'))^{2}-16=4x'^2+16y'^2-16=0$
or, $x'^2+4y'^2=4$
b. In standard form, we have $\frac{x'^2}{4}+\frac{y'^2}{1}=1$, which represents an ellipse.
c. We can graph the equation as shown in the figure.
