If A represents area of ellipse $3x^2+4xy +3y^2=1$ then the value of $\frac{3\sqrt{3}}{\pi}A = ?$ If A represents area of ellipse $3x^2+4xy +3y^2=1$ then the value of $\frac{3\sqrt{3}}{\pi}A = ?$
My approach : 
Since area of ellipse is $\pi ab$ where a is semi major axis and b is semi minor axis. 
Let $(rcos\theta , r sin\theta)$ be any point of the ellipse. Since this ellipse has its centre at (0,0). 
Therefore the given ellipse passes through $(rcos\theta, rsin\theta)$ 
Equation of the ellipse $3x^2+4xy +3y^2=1$ can be written as $3r^2cos^2\theta +4r^2sin\theta cos\theta +3r^2sin^2\theta =1$
$\Rightarrow 3r^2 +r^2 sin2\theta =1$
$\Rightarrow r^2= \frac{1}{3+sin2\theta}$
Please suggest whether this is right way of approaching , and please suggest further. Thanks 
 A: HINT:
As the area is one of the invariants in the Rotation of axes, use this method to eliminate $xy$ to find the values of $a,b$ in the new coordinate Axes 
A: You could use the following fact: If an ellipse is defined implicitly by 
$$\alpha x^2+\beta xy+\gamma y^2=1$$
then its area is given by the following formula
$$A=\frac{2\pi}{\sqrt{4\alpha\gamma-\beta^2}}$$
You could also use the rotation of axis to transform the implicit expression into the traditional formula of the ellipse i.e.
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Then $A=\pi a b$ would be the area of the ellipse. 
A: In general you would need to diagonalize the quadratic form part. But, as you observed, polar coordinates work wonders this time.


*

*$\sin2\theta$ takes all the values in the range $[-1,1]$

*Therefore $3+2\sin2\theta$ takes all the values in the range $[1,5]$.

*Therefore $r^2$ takes all the values in the range _______________

*Therefore $r$ takes all the values in the range _______________

*Therefore the semi-axes are _______ and ________


You fill in the blanks!
