Elliptical cylinder surface area | Analytic geometry

Question

I'm struggling to understand how to solve this kind of problems:

I have to find the equation of a Cylindrical surface area. This area has the generating lines parallel to the axis $z$. The directrix is an ellipse on the floor $Oxy$ with center $C(1;3;0)$ and vertices $A(1;-1;0)$, $B(-1;3;0)$.

---

I tried to make a system with:

• $a = 2$ // As the distance from the center and the $B$ vertex is $2$

• changed $X$ and $Y$ values with those of the vertex $A$ first

• and then with those of the vertex $B$.

  ---

I now have a system of $3$ equations with variables $a$ and $b$, but they're squared only, as they're supposed to be in an ellipse.

I tried to solve it but it doesn't look good as the solution has parameters $x $and $y$ not squared as well.

Solution: $4x^2 + y^2 -x -6y -3 = 0$

Answer 1

A few hints:

• Asking you to find the equation of a surface in three dimensions is something of a red herring. Its generating lines are parallel to the $z$-axis and the directrix ellipse lies entirely in the $x$-$y$ plane, so the equation of the elliptical cylinder will look exactly like the equation of the (two-dimensional) ellipse.
• You’re on the right track by subtracting the center from the two vertices. You should find from this that the axes of the ellipse are parallel to the $x$- and $y$-axes, so in a coordinate system in which the ellipse is at the origin, its equation is $(x'/a)^2+(y'/b)^2=1$.
• To translate the center of the ellipse to a point $(x_c,y_c)$, make the substitutions $x'\to x-x_c$ and $y'\to y-y_c$ in the equation you derived for the ellipse centered at the origin.