Suppose,we have an equation of a parabola $$y=ax^2+bx$$in $xy$ coordinates. We want to find the equation of this parabola in a coordinate which is tilted at an angle ${\theta}$ with the xy such that $x'$ is below $x$. Is there any easy way to do this ?
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Substitute: $$x=x'\cos\theta+y'\sin\theta,\quad y=-x'\sin\theta+y'\cos\theta$$ (clockwise rotation of $\theta$). – Intelligenti pauca Mar 03 '19 at 16:13
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Will it be a parabola in this coordinate also ? – Raihan Amin Mar 03 '19 at 16:27
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Yes, of course: rotation is an isometry. – Intelligenti pauca Mar 03 '19 at 16:30
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But the equ reads : $$ax'^2+by'^2+ax'y'\sin{2\theta}+(b\cos{\theta} +\sin{\theta})x'+(b\sin{\theta}+cos\theta)y'=0$$ .How can i show that it is an equ. of a parabola ? – Raihan Amin Mar 03 '19 at 16:37
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@RaihanAmin Well, how do you define a parabola (or the equation of a parabola)? How to check will depend greatly on that answer. But think about what I said in the other comment: which coordinate grid we put on the plane cannot in any way affect the actual curve. – Arthur Mar 03 '19 at 16:40
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Given a point $(x', y')$ in $x'y'$-coordinates, the $xy$-coordinates of that point is $$ (x'\cos\theta + y'\sin\theta, -x'\sin\theta+y'\cos\theta) $$ A point is on the parabola if the first and second components of the $xy$-coordinates of that point fulfills the given equation. Which is to say, if $$ -x'\sin\theta+y'\cos\theta=a(x'\cos\theta + y'\sin\theta)^2+b(x'\cos\theta + y'\sin\theta) $$
Arthur
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@RaihanAmin Turning the coordinate system a few degrees (or many, for that matter) doesn't change the shape of the curve. – Arthur Mar 03 '19 at 16:33