如何证明圆柱斜切面是椭圆？

方法1，使用解析几何，代入x^2+y^2=1,z=ax+by进行计算。旋转之后得到椭圆方程。
方法2，使用初等集合，找到焦点和准线。
Start with a right circular cylinder intersected at an oblique angle by a plane. We want to show that the intersection is an ellipse. The first step is to construct two spheres, each with radius equal to the radius of the cylinder and center on the cylinder axis, so they will both be tangent to the cylinder. The centers of the spheres are located on the cylinder axis such that each sphere will be tangent to the intersecting plane, one above it and one below it. (This is the first construction the details of which I'll omit. It should be clear that such spheres can be defined and are unique.) Now we consider a point P on the intersection curve of the plane and the cylinder. Construct a line from P to A, the point of tangency between the upper sphere and the plane. Next construct a line from P, parallel to the axis of the cylinder, and extend it to the circular intersection of the upper sphere and the cylinder. Call the point of intersection between this line and the circle U. Lines PA and PU are both tangent lines drawn from a point external to a sphere; therefore they are equal in length. Next, construct similar lines PB and PL, from P to the point of tangency of the lower sphere and the plane, and from P to the circular intersection of the lower sphere and the cylinder, respectively. Again, we have PB=PL. Now, since PU+PL is simply the distance between the circular intersections of the two spheres and the cylinder, this sum will be the same regardless of where the point P is chosen on the intersection of the plane and the cylinder. Since the PA=PU and PB=PL, however, we have also that PA+PB will be equal to PU+PL, and thus it, too, will be the same regardless of where the point P is chose. Thus the intersection is an ellipse with foci at A and B. Thanks to tiny-tim for pointing out that similar proofs can be constructed for the intersections of a plane and a right circular cone (all three cases, ellipse, parabola, and hyperbola). In these cases the circular intersections of the spheres and the cone will not be great circles, as they are for the cylinder, but this turns out not to matter. The case of the parabola is also somewhat different, since there will be only one sphere. In that case, the latus rectum of the parabola is constructed by finding the intersection of the first plane and the plane containing the circular intersection of the single sphere and the cone.

A代表圆柱体斜切后形成的椭圆

当然用椭圆的性质了