Construct equation of ellipse, if focuses are $F_1 (-1, -1), F_2 (1, 1)$, height (perpendicular to $F_1 F_2$) $b = 1$ First of all, I know the right answer. It's $2x^2 - 2xy + 2y^2 - 3 = 0$.
What is unknown to me, is how to obtain that answer.
I've tried to make equation like $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
We know $b$.
$c$ is half-distance between $F_1$ and $F_2$, so $c = \sqrt{2}$  (full distance $ = \sqrt{8} = 2\sqrt{2}$)
According to some formulas, $b = \sqrt{a^2 - c^2}$, hence $a^2 = b^2 + c^2 = 1^2 + (\sqrt{2})^2 = 3$
So equation "should be" the next one: $\frac{x^2}{3} + \frac{y^2}{1} = 1$
But it's a wrong equation. Obviously, this ellipse has been rotated by 45 degrees. Some posts hinted me that I should put $(a_0 x + a_1 y)^2$ in numerator.
Unfortunately, I don't know how to "rotate" ellipse in such form of equation. Perhaps, I need somehow to come up with more generalized form of equation
$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, but still I don't know to identify the coefficients
Anyway, I would be glad to perceive any help on your behalf.
 A: If $\left(\begin{matrix}X\\Y\end{matrix}\right)$ is the image of the vector $\left(\begin{matrix}x\\y\end{matrix}\right)$ after rotation by $45^o$, then $$\left(\begin{matrix}X\\Y\end{matrix}\right)=M\left(\begin{matrix}x\\y\end{matrix}\right)$$
Where $$M=\left(\begin{matrix}\cos 45&-\sin 45\\\sin 45&\cos 45\end{matrix}\right)$$
$$\implies \left(\begin{matrix}x\\y\end{matrix}\right)=M^{-1}\left(\begin{matrix}X\\Y\end{matrix}\right)$$
Where $$M^{-1}=\left(\begin{matrix}\cos 45&\sin 45\\-\sin 45&\cos 45\end{matrix}\right)$$
Hence $$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{matrix}\right)\left(\begin{matrix}X\\Y\end{matrix}\right)$$
So, taking the equation of the ellipse you found, i.e. $x^2+3y^2=3$, substitute into this $$x=\frac{1}{\sqrt{2}}X+ \frac{1}{\sqrt{2}}Y$$ and $$y=-\frac{1}{\sqrt{2}}X+ \frac{1}{\sqrt{2}}Y$$ and you get the equation of the required rotated ellipse $$2X^2-2XY+2Y^2=3$$
