termegolz6

Answered

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

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

Answer & Explanation

hottchevymanzm

Expert

2022-07-20Added 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 $({x}^{\prime}/a{)}^{2}+({y}^{\prime}/b{)}^{2}=1$.

- To translate the center of the ellipse to a point $({x}_{c},{y}_{c})$, 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.

- 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 $({x}^{\prime}/a{)}^{2}+({y}^{\prime}/b{)}^{2}=1$.

- To translate the center of the ellipse to a point $({x}_{c},{y}_{c})$, 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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