2

Question: Parametric equation of surface S is given below: $$x=6\cos (t+\frac\pi 3).y=3\sin t,z=u \text{ and } 0\leq t\leq 2\pi .u\in R$$

Find the curve which is intersecting with surface S's plane: $y={x\over\sqrt 3}$ and explain what type of intersection it is.

I can't seem to find a solution, can you help me, I mean can you show me a way to solve this question?

Matthers
  • 33
  • 5
  • 1
    It's the "cylinder" over the ellipse $x^2+2\sqrt{3}xy+4y^2-9=0.$ The plane $y=\frac{x}{\sqrt{3}}$ will cut two line segments on either side, over the intersection points of the line and conic in the xy-plane. – Jan-Magnus Økland Jun 26 '20 at 12:38

0 Answers0