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I'm trying to find the graphic of the curve of the intersection of the hyperbolic paraboloid $z=y^2-x^2$ and the cylinder $x^2+y^2=1$. I could show them together:

I want to see the curve of intersection highlighted, is it possible? I've already searched everywhere without success.

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marked as duplicate by Szabolcs, Community Mar 30 '17 at 14:11

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  • $\begingroup$ Have you seen this? $\endgroup$ – J. M. will be back soon Mar 30 '17 at 13:27
  • $\begingroup$ @J.M. Thank you very much. I followed one of the answers and plot: ContourPlot3D[{x^2 + y^2 == 1, z=y^2-x^2}, {x, -4, 4}, {y, -4, 4}, {z, -2, 2},Contours -> {0}, ContourStyle -> Opacity[0], Mesh -> None, BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> {{Green, Tube[.03]}}}, Boxed -> False]. Mathematica tells me y^2-x^2 is not a valid variable $\endgroup$ – user26832 Mar 30 '17 at 13:38
  • $\begingroup$ Clear all variables (ClearAll["Global`*"]) and try again. Also, your second equation should be z == y^2 - x^2; = and == are different things. $\endgroup$ – J. M. will be back soon Mar 30 '17 at 13:40
  • $\begingroup$ @J.M. Thank you again, it worked. However I couldn't put them together, Mathematica says: Show :Could not combine the graphics objects in Showp1,p2, . General :Further output of Get::noopen will be suppressed during this calculation. Get :Cannot open 1241>> . Get :Cannot open 1241>> . Get :Cannot open 1>> . $\endgroup$ – user26832 Mar 30 '17 at 13:45
  • $\begingroup$ @J.M. I put Show[p1,p2,p3], where p1=Plot3D[z=y^2-x^2,{x,-2,2},{y,-2,2}], p2=ContourPlot3D[x^2 + y^2 == 1, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}],p3=ContourPlot3D[{x^2 + y^2 == 1, z==y^2-x^2}, {x, -4, 4}, {y, -4, 4}, {z, -2, 2},Contours -> {0}, ContourStyle -> Opacity[0], Mesh -> None, BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> {{Green, Tube[.03]}}}, Boxed -> False] $\endgroup$ – user26832 Mar 30 '17 at 13:46

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