To find the angle of rotation of a conic section, we have fisrs to see the following form of the equation:
[tex]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/tex]
Given thah form, we can find the angle of rotation by the equation:
[tex]\tan 2\theta=\frac{B}{A-C}[/tex]
From first the given section:
[tex]17x^2+32xy-7y^2=75[/tex]
We get that A = 17, B = 32 and C = -7. So,
[tex]\begin{gathered} \tan 2\theta=\frac{32}{17-(-7)}=\frac{32}{24}=\frac{4}{3} \\ 2\theta=\arctan (\frac{4}{3})=53.1\degree \\ \theta=26.6\degree \end{gathered}[/tex]
The angle of the first given section is 26.6°.
Now, for the second one:
[tex]32x^2+50xy+7y^2=100[/tex]
We have A = 32, B = 50 and C = 7, so:
[tex]\begin{gathered} \tan 2\theta=\frac{50}{32-7}=\frac{50}{25}=2 \\ 2\theta=\arctan 2=63.4\degree \\ \theta=32.7\degree \end{gathered}[/tex]
The angle of the second given section is 32.7°.
