The question is:
If A represents the area of the ellipse $\,3x^2+4xy+3y^2=1$, then the value of $\frac{3\sqrt5}{\pi}A$ is
For this I used rotation of axes for eliminating the $xy$ term from the equation so that I can get the equation in the standard from.
Rotating by $\theta$, we get,
$$x=X\cos\theta-Y\sin\theta$$$$y=X\sin\theta+Y\cos\theta$$
So, equation of ellipse now becomes,
$$3(X\cos\theta-Y\sin\theta)^2+4(X\cos\theta-Y\sin\theta)(X\sin\theta+Y\cos\theta)+3(X\sin\theta+Y\cos\theta)^2=1$$
On simplifying, gives, $$6X^2+6Y^2+4\sin\theta\cos\theta(X^2-Y^2)+4XY(\cos^2\theta-\sin^2\theta)=1$$
Making coefficient of $XY$ zero, we have, $$4(\cos^2\theta-\sin^2\theta)=0$$ $$\implies \theta=\frac{\pi}{4}$$
Putting the value of $\theta$ in the equation, we get, $$8X^2+4Y^2=1$$
So,
Semi-major axis$=\frac{1}{2\sqrt2}$
Semi-minor axis$=\frac12$
So, $$A=\pi\times\frac{1}{2\sqrt2}\times\frac12=\frac{\pi}{4\sqrt2}$$
Then, $\frac{3\sqrt5}{\pi}A=\frac{3\sqrt5}{4\sqrt2}$ but the answer is $3$.
I don't know where I went wrong. Can anybody check my calculations or is there better way to do this?