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? 

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.
