Answer
a. $x'^2+2y'^2-1=0$
b. $\frac{x'^2}{1}+\frac{y'^2}{1/2}=1$
c. see graph
Work Step by Step
a. Step 1. Given $34x^{2}-24xy+41y^{2}-25=0$, we have $A=34, B=-24, C=41$, and the rotation angle
$cot(2\theta)=\frac{A-C}{B}=\frac{34-41}{-24}=\frac{7}{24}$,
Step 2. We have the transformations (see also exercise 17):
$x=\frac{1}{5}(4x'-3y')$ and $y=\frac{1}{5}(3x'+4y')$
Step 3. The equation becomes
$34x^{2}-24xy+41y^{2}-25=34(\frac{1}{5}(4x'-3y'))^{2}-24(\frac{1}{5}(4x'-3y'))(\frac{1}{5}(3x'+4y'))+41(\frac{1}{5}(3x'+4y'))^{2}-25=25x'^2+50y'^2-25=0$
or, $x'^2+2y'^2-1=0$
b. In standard form, we have
$\frac{x'^2}{1}+\frac{y'^2}{1/2}=1$
which represents an ellipse.
c. We can graph the equation as shown in the figure.
