Hi my name is Matthew and I am a uni student, I am stuck on a question, i hope you can help: Let C in R^{3} be the cone defined by x^{2} + y^{2}  z^{2} = 0 (A) Let P be the plane described by x + 2z = 1 (i) Find a description of P in terms of two parameters s and t (ii) Substitute your formulas for (x, y, z) in P in terms of s and t into the equation defining C. This gives a relation involving s and t which implicity describes the intersection of C and P. (iii) Sketch this intersection set in the stplane. (B) Suppose now that the plane P is described by x + az = b. Find the conditions on a, b such that the intersection of C and P is (i) an ellipse (ii) a parabola (iii) a hyperbola. Thankyou for your help. Matthew Hi Matthew, You can attack this two ways. Ideally, you would do both and see that they match up. One way (the way the questions posed seem to suggest) is analytically, with formulae. I will NOT give those details. However, here is a sample. If the plane is horizonatal z= c, then the intersection is found by replacing z by c, and seeing the answer as a plane curve: x^{2} + y^{2} c^{2} = 0 or x^{2} + y^{2} = c^{2} A nice circle, centered at the origin. You can investigate some other versions (z = ax + b) and find that this is the only situation where it remains a circle. etc. The second way is to 'see it' in 3space. This cone is what they call a right circular cone. This means (as the algebra above indicated) that horizontal slices are circles. Now try moving various planes around and seeing how they slice the cone, extended in two directions (positive and negative z). If you take a plane through the origin, either:
I you now take this tangent plane and translate it away from the origin, it will only hit one side of the cone  in a parabola. If you take something like a vertical plane (x=0), through the origin, you get two intersecting lines. If you translate this (x=c) you move out and get a hyperbola, with the original intersecting lines Still visible as the assymptotes of the hyperbola. One can do more, but I think this is enough from me. Walter
