<s>[INST] 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!
 [/INST] 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.   $V =  \left(-\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2} \right)$ 
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