       Re: Graphics << Implicit vs ContourPlot

• To: mathgroup at smc.vnet.net
• Subject: [mg122756] Re: Graphics << Implicit vs ContourPlot
• From: Patrick Scheibe <pscheibe at trm.uni-leipzig.de>
• Date: Thu, 10 Nov 2011 06:49:29 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201111091124.GAA11084@smc.vnet.net>

```Hi,

just say Axes->True and Frame->False:

ContourPlot[{7 x^2 - 6 Sqrt x y + 13 y^2 - 16 ==
0, ((x^2)/2^2) + ((y^2)/2^2) == 1, y == x, y == -x, y == 0,
x == 0}, {x, -3, 3}, {y, -3, 3}, AspectRatio -> 1, Axes -> True,
Frame -> False]

Cheers
Patrick

On Wed, 2011-11-09 at 06:24 -0500, John Accardi wrote:
> Goal: Show students a plot of both an ellipse with x axis as ellipse's
> transverse axis
> and the same ellipse in an x'y' coordinate plane that is rotated some angle
> with respect the the original xy coordinate plane.  (All in one plot)
>
> I used:
>
> << Graphics`ImplicitPlot`; ImplicitPlot[{7 x^2 - 6 Sqrt x y +
>      13 y^2 - 16 == 0, ((x^2)/2^2) + ((y^2)/2^2) == 1, y == x,
>    y == -x}, {x, -3, 3}, AspectRatio -> 1.25]
>
> which works well but I had to hard fix the axis of rotation at 45 degrees
> and plot it (y == x and y == -x).  I also get an obsolete warning and
> the suggestion to use the new ContourPlot for this in the future:
>
> General::obspkg: "\!\(\"Graphics`ImplicitPlot`\"\) is now obsolete.
> The legacy version being loaded may conflict with current Mathematica
> functionality. See the Compatibility Guide for updating information."
>
> So I try to accomplish the same graph with ContourPlot:
>
> ContourPlot[{7 x^2 - 6 Sqrt x y + 13 y^2 - 16 ==
>     0, ((x^2)/2^2) + ((y^2)/2^2) == 1, y == x, y == -x, y == 0,
>    x == 0}, {x, -3, 3}, {y, -3, 3}, AspectRatio -> 1.25]
>
> which gets me close but I have lost traditional plotting of the xy axes
> (no tick marks).  Instead I get ContourPlots boxed style coordinate system.
>
> Question: How can I get my old style axes back in the context of
> ContourPlot?
>
> Thank you for any insights.
>

```

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