# Parabola in a tilted coordinates

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 ?

• Substitute: $$x=x'\cos\theta+y'\sin\theta,\quad y=-x'\sin\theta+y'\cos\theta$$ (clockwise rotation of $\theta$). – Aretino Mar 3 at 16:13
• Will it be a parabola in this coordinate also ? – Raihan Amin Mar 3 at 16:27
• Yes, of course: rotation is an isometry. – Aretino Mar 3 at 16:30
• 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 3 at 16:37
• @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 3 at 16:40

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 $$-x'\sin\theta+y'\cos\theta=a(x'\cos\theta + y'\sin\theta)^2+b(x'\cos\theta + y'\sin\theta)$$