Answer
a. $x'-2y'^2=0$
b. $y'^2=\frac{1}{2}x'$
c. see graph
Work Step by Step
a. Step 1. Given
$32x^{2}-48xy+18y^{2}-15x-20y=0$
we have
$A=32, B=-48, C=18$
The rotation angle is
$cot(2\theta)=\frac{A-C}{B}=\frac{32-18}{-48}=-\frac{7}{24}$,
Step 2. We have the transformations (see also exercise 21):
$x=\frac{1}{5}(3x'-4y')$ and $y=\frac{1}{5}(4x'+3y')$
Step 3. The equation becomes
$32x^{2}-48xy+18y^{2}-15x-20y=32(\frac{1}{5}(3x'-4y'))^{2}-48(\frac{1}{5}(3x'-4y'))(\frac{1}{5}(4x'+3y'))+18(\frac{1}{5}(4x'+3y'))^{2}-15(\frac{1}{5}(3x'-4y'))-20(\frac{1}{5}(4x'+3y'))=-25x'+50y'^2=0$
or, $x'-2y'^2=0$
b. In standard form, we have $y'^2=\frac{1}{2}x'$ which represents a parabola.
c. We can graph the equation as shown in the figure.
