Finding the vertex and focus of a rotated parabola

Question

So I begun with the following equation : $x^2+2xy+y^2+2\sqrt{2}x-2\sqrt{2}y+4=0$

I transformed it in the following : $y'=\frac{x'^2}{2}+1$

I had to do a rotation of $\frac{\pi}{4}$ of the xy axis. (counter clock wise) My question is how do you find the focus and vertex of the rotated parabola ? I think that the vertex is going to be (0,1) but I'm not sure if this is the case when we rotate it.

Thank you!

Answer 1

The transform equations to use for rotating any curve are $$ x' = (x \cos θ - y \sin θ)$$ $$y' =(x \sin θ + y \cos θ)$$ (These basically rotate the axes, but when we view them with static axes the graph rotates)

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In this case, we get $$x' = \frac{x+y}{\sqrt{2}}$$ $$y' = \frac{y-x}{\sqrt{2}}$$ $$$$$$$$ $$x^2+2xy+y^2+2\sqrt{2}x-2\sqrt{2}y+4=0$$ After the transformation we get: $$\Rightarrow \frac{1}{2}(x+y)^2 + (x+y)(y-x) + \frac{1}{2}(y-x)^2 + 2(x+y) - 2(y-x)+4=0$$ $$\Rightarrow x = -\frac{y^2}{2}-1$$

This is just a horizontal parabola scaled down and shifted left by $1$; therefore, the vertex is at $(-1,0)$

Answer 2

Assuming you have done the coordinate transformation correctly, then the basic idea is that you calculate the vertex and focus of the transformed parabola, then perform the inverse transformation on those coordinates to recover the vertex and focus in the untransformed (original) coordinates.

So for example, if your transformation constituted first a translation of the form $$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} x - h \\ y - k \end{bmatrix},$$ followed by a counterclockwise rotation of $\theta$ $$\begin{bmatrix} x'' \\ y'' \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} x' \\ y' \end{bmatrix},$$ and the resulting parabola had vertex at $(x'', y'') = (p'',q'')$, then you'd apply the inverse of the rotation matrix to $(p'', q'')$, and then the inverse of the translation, giving $$\begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} p'' \\ q'' \end{bmatrix} + \begin{bmatrix}h \\ k \end{bmatrix}.$$ The same principle applies to the focus.

Answer 3

I’ll do the same thing as @heropup, but without the notation. To save myself typing, I’ll set $c=1/\sqrt2=\cos 45^\circ=\sin45^\circ$. Then you want a rotation of $45^\circ$, and to do this I’ll set \begin{align} x&=cX-cY\\y&=cX+cY\,. \end{align} Make these substitutions, and if I’m not mistaken, you get a nice equation of form $Y=\alpha X^2+\beta$, but I’ll leave it to you to find these numbers. You get points that are the $(X,Y)$-coordinates of your vertex and focus (you know how to do that), and plug these into the displayed formulas to get the $(x,y)$-coordinates. (PS: make sure you express $2\sqrt2$ correctly as $4c$. You’ll see that the letter $c$ drops out completely.)