conic sectionsgeometry
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!
To find the vertex of that parabola, you can make use of a nice general result:
given three tangents of a parabola, if $AB$ is that part of a tangent which is comprised between the other two tangents (see diagram), then the projection of $AB$ onto a line perpendicular to the axis has a fixed length, independent of the position of $AB$.
In our particular case, if $V$ (vertex) and $D$ are the tangency points of the "outer" tangents, $C$ is their intersection point, and $B'$, $D'$ are the projections of $B$, $D$ on tangent $VC$ (which is perpendicular to the axis), we obtain from the above result:
$$AB'=CD'=VC.$$
As points $A$, $B$, $C$, $B'$ can be readily found from the data, one can also easily find vertex $V$ and tangency point $D$.
A nice property of the parabola states that: the perpendicular from the focus to any tangent intersects it, and the tangent through the vertex, at the same point. Hence, if the tangent at vertex $V$ intersect the axes at $A'=(\alpha,0)$ and $B'=(0,\beta)$, then focus $F$ has coordinates $(\alpha,\beta)$.
The axis $FV$ is perpendicular to $A'B'$, hence its slope is $\alpha/\beta$. But the tangent at $A$ forms equal angles with the axis and $FA$, hence the slope of line $FA$ is $-\alpha/\beta$. The same goes for line $FB$, hence we have the equations: $$ {\beta-b\over\alpha}={\beta\over\alpha-a}=-{\alpha\over\beta}, $$ which can be solved to find: $$ \alpha={ab^2\over a^2+b^2},\quad \beta={a^2b\over a^2+b^2}. $$ The directrix is the parallel to $A'B'$ passing through the origin.
Best Answer
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 $(-1,0)$