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I want to scale transform the equation (not the graph) of a circle. Specifically, I want to scale 4x by the x-axis and 3x by the y-axis. I use the following x^2 + y^2 == 1 /. Thread[{x, y} -> ScalingTransform[{4, 3}][{x, y}]] The result is 16x^2 + 9y^2 == 1 and it is wrong. How to get the correct result of (x^2)/16 + (y^2)/9 == 1?

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  • $\begingroup$ Welcome to Mathematica SE. To get started:1) take the introductory tour now,2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge,3) remember to accept the answer, if any, that solves your problem, by clicking checkmark sign,4) give help too, by answering questions in your areas of expertise. $\endgroup$
    – user49048
    Mar 10, 2022 at 0:02
  • $\begingroup$ Please note that transform the equation ` x^2+y^2-1==0` is the opposite of the transform of the mapping f[x_,y_]=x^2+y^2-1. $\endgroup$
    – cvgmt
    Mar 11, 2022 at 1:33

2 Answers 2

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The simple

x^2 + y^2 == 1 /. 
 Thread[{x, y} -> ScalingTransform[{1/4, 1/3}][{x, y}]]

yields

x^2/16 + y^2/9 == 1

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  • $\begingroup$ This easy fix loses generality. If I want to rotate the parabola for 60 degree, the following y==x^2+2x-3 /. Thread[{x, y} -> RotationTransform[Pi/3, {0,0}][{x, y}]] gives the equation after -60 degree rotation. Is there a more general way to fix the rotation and scaling problem and produce the correct equation, instead of the opposite equation? $\endgroup$
    – Richard
    Mar 10, 2022 at 2:00
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    $\begingroup$ @Richard with all due respect, the question of the OP is fully answered in my opinion. ScalingTransform was reported to cause some unwanted behavior and I provided an answer to that. I could NOT have guessed what you mentioned in your comment as you understand. If you are interested in RotationTransform perhaps you could update the OP and describe that as well in a nice context -as you did for the original question- or start a new thread. I think that this is good practice for other users as well. $\endgroup$
    – user49048
    Mar 10, 2022 at 2:04
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    $\begingroup$ @Richard I'd venture the guess that -Pi/3 would give the intended result. In general it seems that taking the inverse of transformation should give you what you want. $\endgroup$
    – infinitezero
    Mar 10, 2022 at 11:16
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I think you're mixing up active and passive transformations. It looks to me that Mathematica's definition for Thread[{x, y} ->ScalingTransform[{4, 3}][{x, y}]] is to replace x with 4x and y with 3y, which is an active transformation.

In your original post, you say that you want to "scale 4x by the x-axis and 3x by the y-axis," which I believe you are conceptualising as a passive transformation. The active transformation which achieves this effect is to replace x with x/4 and y with y/3, as described in @kcr's answer. This change of perspective also solves your problem with RotationTransform in the comments.

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