Finding the vertex and focus of a rotated parabola 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!
 A: 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)    

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$$
This is just a horizontal parabola scaled down and shifted left by $1$; therefore, the vertex is at $(-1,0)$
A: Assuming you have done the coordinate transformation correctly, then the basic idea is that you calculate the vertex and focus of the transformed parabola, then perform the inverse transformation on those coordinates to recover the vertex and focus in the untransformed (original) coordinates.
So for example, if your transformation constituted first a translation of the form $$\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.
A: 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.)
A: I'm going to assume the OP wanted the vertex and focus of the original tilted parabola since they already had rotated it to a standard form where those values are easier to find.
Rather than rotating, its convenient  to do everything in place since we're looking for the focus and by definition: for any point on the parabola the distance to the focus is equal to the distance to the directrix.  
First let the focus F = (a,b) and  the directrix is a linear equation  y = mx + c
We need to find the point P (p, mp + c) on the directrix such that the segment between it
and an arbitrary point on the parabola is perpendicular to the directrix i.e. $-\dfrac{1} {m} = \dfrac{(y - mp - c)}{x - p}$
Solving for p we get $p = \dfrac{1}{1 + m^2}(my + x- mc)$  and $P = \left(\dfrac{1}{1 + m^2}(my + x- mc),  \dfrac{m}{1 + m^2}(my + x - mc) + c \right)$
Then our general equation for the parabola is: 
$(x-a)^2 + (y-b)^2 = (x - \dfrac{1}{1 + m^2}(my + x- mc))^2 + (y - \dfrac{m}{1 + m^2}(my + x - mc) - c)^2$ 
This simplifies nicely to $(x-a)^2 + (y-b)^2  =  \dfrac{1}{1 + m^2}(mx - y + c)^2$
If we move everything to the lhs we get this form:
$x^2 + m^2y - 2mxy + (-2(m^2 + 1)a -2c)x + (-2(m^2 + 1)b + 2c)y + ((m^2 + 1)(a^2 + b^2
) - c^2) = 0$
So if the coefficient of $x^2$ in the original equation is 1 we can easily find m.
This is already the case in our parabola so  here m = 1.
Now we're left with $(x-a)^2  + (y - b^2) = \dfrac{1}{2}(x - y + c)^2$
Corresponding with the original equation gives us a system of equations:


*

*$2a^2 + 2b^2 - c^2 = 4$

*$-4a -2c = 2 \sqrt{2}$

*$-4b +2c = -2 \sqrt{2}$
Solving for c and then substituting back in to find a and b we get:
$(x + \frac{3}{4}\sqrt{2})^2 + (y  - \frac{3}{4}\sqrt{2})^2 = \frac{1}{2}(x - y + \frac{\sqrt{2}}{2})^2$
The first part is done.  The focus $F = \left( -\dfrac{3}{4}\sqrt{2}, \dfrac{3}{4}\sqrt{2} \right)$
The vertex is halfway between F and the directrix. So first we look at the axis of symmetry y = -x + d which goes through F.  Solving d = 0.  This intersects the directrix where $-x  = x +  \frac{\sqrt{2}}{2}$  at I = $\left(-\dfrac{1}{4}\sqrt{2}, \dfrac{1}{4}\sqrt{2}\right)$ 
And the vertex V is at the midpoint of the segment between F and I.   $V =  \left(-\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2} \right)$ 
