Quadric Surfaces — Question 10

Question 10 – Solution

Strategy Express $x^2+y^2$ and $xy$ in rotated coordinates so the cross term disappears.

Step 1: Invert and substitute We have $$x=\frac{u+v}{\sqrt 2},\quad y=\frac{u-v}{\sqrt 2},\quad x^2+y^2=u^2+v^2,\quad xy=\frac{u^2-v^2}{2}.$$ Therefore $$5(u^2+v^2)-3(u^2-v^2)-z^2=0, \quad\boxed{2u^2+8v^2-z^2=0}.$$

Step 2: Classify This is an elliptic cone with vertex at the origin and axis the $z$-axis.

Step 3: Traces For $z=c\ne 0$, $$\boxed{\frac{u^2}{c^2/2}+\frac{v^2}{c^2/8}=1}.$$ For $v=0$, $z=\pm\sqrt 2u$; for $u=0$, $z=\pm 2\sqrt 2v$. At $z=0$, only the vertex remains.

Step 4: Interpret rotation The $u$-axis is $x=y$ and the $v$-axis is $x=-y$, so the principal horizontal directions are rotated $45^\circ$ from the original axes.