Timeline for Drawing parabola in Graphics
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 23, 2017 at 14:35 | comment | added | Kuba | I think you can assume someone wants to plot a parabola based on a/b/c parameters. | |
| Nov 23, 2017 at 14:03 | comment | added | Sumit |
The only problem is, there is no specification for the parabola. Otherwise, appearance can be deceived, at least for a certain range. For example ContourPlot[{x^2/1 + (y - 3)^2/9 == 1, x^2 == 2/3 y}, {x, -1.5, 1.5}, {y, 0, 3}] and ContourPlot[{x^2/1 + (y - 3)^2/9 == 1, x^2 == 2/3 y}, {x, -1.5, 1.5}, {y, 0, 0.5}]
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| Nov 23, 2017 at 13:25 | comment | added | Kuba |
It can be saved but you need to get math straight, start with: Show[ Graphics[{Circle[{0, 1000 }, {10, 1000}, {Pi, 2 Pi }]}, PlotRange -> 1], Plot[5 (x + .1)^2, {x, -1, 1}] , Frame -> True ] parameters for circle are out of the blue but you can try to provide approximate formula.
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| Nov 23, 2017 at 11:06 | history | edited | Sumit | CC BY-SA 3.0 |
added 9 characters in body
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| Nov 23, 2017 at 11:06 | comment | added | Sumit | @Kuba, you are right and I am a disaster again. It is indeed (half) an ellipse. My high school math marks still make sense. | |
| Nov 23, 2017 at 9:59 | comment | added | Kuba |
While parabola is a special case of ellipse I don't think Circle syntax allows this. Or did I miss the point?
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| Nov 23, 2017 at 9:53 | history | answered | Sumit | CC BY-SA 3.0 |