 termegolz6

2022-07-19

Elliptical cylinder surface area | Analytic geometry
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 xand y not squared as well.
Solution: $4{x}^{2}+{y}^{2}-x-6y-3=0$ hottchevymanzm

Expert

Explanation:
- 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 $\left({x}^{\prime }/a{\right)}^{2}+\left({y}^{\prime }/b{\right)}^{2}=1$.
- To translate the center of the ellipse to a point $\left({x}_{c},{y}_{c}\right)$, make the substitutions ${x}^{\prime }\to x-{x}_{c}$ and ${y}^{\prime }\to y-{y}_{c}$ in the equation you derived for the ellipse centered at the origin.

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