Find the equation of the cylindrical surface whose…

Find the equation of the cylindrical surface whose generators are parallel to the vector $(1,1,2)$ and whose directrix is the curve \[ x+y+z=0, \qquad 5x^2+6xy+5y^2=4. \]

Step 1 · General solution

From the previous problem, every cylinder with generators parallel to $(1,1,2)$ has the form \[ z = 2x + \Phi(x-y). \tag{$*$} \]

Step 2 · Apply the directrix condition

From $x+y+z=0$: $z=-(x+y)$. Substituting into ($*$): \[ \Phi(x-y) = -3x-y. \] With $t=x-y$ (so $x=y+t$), the second directrix equation $5x^2+6xy+5y^2=4$ becomes $16y^2+16ty+5t^2=4$, a quadratic in $y$ with roots $y=\dfrac{-2t\pm\sqrt{4-t^2}}{4}$. Then \[ -3x-y = -4y-3t = -t\mp\sqrt{4-t^2}, \] so $(\Phi(t)+t)^2=4-t^2$.

Step 3 · Equation of the cylinder

Substituting $t=x-y$ and $\Phi(t)=z-2x$ into $(\Phi(t)+t)^2=4-t^2$: \[ (z-2x+(x-y))^2 = 4-(x-y)^2 \;\Longrightarrow\; (z-x-y)^2+(x-y)^2=4. \] Expanding both squares: \[ 2x^2+2y^2+z^2-2xz-2yz = 4. \]

Verification

On $z=-(x+y)$: $2x^2+2y^2+(x+y)^2+2x(x+y)+2y(x+y)=2x^2+2y^2+3(x+y)^2=5x^2+6xy+5y^2=4$. $\checkmark$
For $G=2x^2+2y^2+z^2-2xz-2yz-4$: $G_x=4x-2z$, $G_y=4y-2z$, $G_z=2z-2x-2y$, so \[ p+q = \frac{-(4x-2z)-(4y-2z)}{2z-2x-2y} = \frac{4(z-x-y)}{2(z-x-y)}=2. \checkmark \]