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).

termegolz6

termegolz6

Answered question

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 6 y 3 = 0

Answer & Explanation

hottchevymanzm

hottchevymanzm

Beginner2022-07-20Added 15 answers

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 ( 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 x x c and y y y c in the equation you derived for the ellipse centered at the origin.

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