<s>[INST] Area of ellipse which is not in standard form By graphing device i understand that $x^2+xy+y^2=1$ is ellipse. By some geometry i find area of above ellipse which comes out $\pi$ (is it right?), but it was easy case. Is there any quick method or standard formula to calculate it or we have to convert it into standard form always and then calculate? 
 [/INST] If the conic $ Q(x,y) = ax^2 + by^2 + c + 2hxy + 2fy + 2gx$ is an ellipse, then if we translate the orgin of our coordinate system to the centre of $ Q$ to get,
$$ {R(X,Y) = Q(X + u, Y + v)} = aX^2 + b Y^2 + 2hXY + c^\prime$$
where $(u, v)$ is the centre of the conic given. We need to find $(u, v)$ and $c^\prime$  (practically we only need $c'$). 
Anyway $(u, v)$ is the solution of $$\begin{cases}\dfrac{\partial Q}{\partial x}  = 0 \\ \dfrac{\partial Q}{\partial y}  = 0 \end{cases}$$ 
and $c^\prime$ is given by $ c^\prime =\dfrac{\Delta}{\delta}$ where $$\delta = \begin{vmatrix} a & h \\ h & b\end{vmatrix}.$$ and $$\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}.$$
The problem reduces to finding the area of the curve $R(X, Y) = 0$, which is equivalent to finding the area of $ S (X, Y) = \dfrac{-a}{c^\prime}X^2 + \dfrac{-b}{c^\prime} Y^2 + 2\dfrac{-h}{c^\prime}XY - 1$.
Let $\alpha = \dfrac{-a}{c^\prime}
, \beta = \dfrac{-b}{c^\prime}$ and $\gamma  = \dfrac{-h}{c^\prime}$, then we have $ S(X, Y) = \alpha X^2 + \beta Y^2 + 2\gamma XY - 1$. 
To write $ S$ in standard form we need to eliminate $ XY$ term. If $ \gamma = h \ne 0$ then that means the ellipse is rotated by some angle $\theta$ given by the relation, $$\tan (2\theta) = \dfrac{2\gamma}{ \alpha - \beta} = \dfrac{2h}{a - b}.$$
So to remove $XY$ term we rotate $S$ by $-\theta$, $${X \choose Y} = \left[\begin{matrix}\cos\theta & -\sin \theta \\ \sin \theta& \cos \theta \end{matrix}\right] {X^\prime \choose Y^\prime}.$$
We will then get,
$$ S(X^\prime, Y^\prime)  = \dfrac{X{{^\prime}^2} }{\alpha'^2} + \dfrac{Y{{^\prime}^2} }{\beta'^2} - 1 = 0$$ 
The area of which is given by $\pi \alpha' \beta'$.

For example, Let $Q(x,y) = x^2 + 2y^2 + 2xy + 4x + 4y + 1$,
Here we have $c^\prime = -3$ and $\theta = \dfrac12 \tan^{-1} (-2)$.
First translating the origin we get, $S(X, Y) = \dfrac{1}{3} X^2 + \dfrac23 Y^2 + 2\dfrac13XY  - 1$. 
Now we rotate the cooridinate system by $\dfrac12 \tan^{-1} (2)$ to get the equation of the ellipse as,
$$\dfrac{X'^2}{\dfrac6{3- \sqrt{5}}} + \dfrac{Y'^2}{\dfrac6{3+ \sqrt{5}}} = 1 $$
So the area is $3\pi$.
Not the easiest method but the simplest method for sure.
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