<s>[INST] Find area of ellipse $5x^2 -6xy +5y^2=8$ 
Find the  area of ellipse whose  equation in the  $xy$- plane  is given by  $5x^2 -6xy +5y^2=8$

My attempt  : I know  that area   of ellipse $ = \pi a b$ ,where $a$ is  semi-major axis and $b$ is  semi minor axis
Now  if we make  matrix $\begin{bmatrix} 5 & -3 \\-3& 5\end{bmatrix}$ Here  eigenvalue  $\lambda_1=  8 ,\lambda_2=2$
That  is area of ellipse$ = \pi \frac{1}{\sqrt\lambda_2} \frac{1}{\sqrt\lambda_1}= \pi \frac{1}{2\sqrt 2}\frac{1}{\sqrt2}$
Is its  correct ?
 [/INST] Center of ellipses is $(0,0)$ and any point on the chord passing through center is $(r\cos{\theta} , r\sin{\theta} )$ . Let chord intersect ellipses at points $A$ and $B$ . Put the point in equation of ellipses we get 
$$(5\cos^2{\theta})r^2-(6\cos{\theta} \sin{\theta} ) r^2 + (5\sin^2{\theta} )r^2 =8$$
$$(5-3\sin^2{\theta})r^2=8$$
$$r^2=\frac{8}{5-3\sin^2{\theta} }$$
Maximum $r = a $ and minimum $r=b$ and maximum $r$  we get when $\sin^2{\theta} = 1$ and minimum $r$ we get when $\sin^2{\theta}=-1$ .
Hence $$a^2=\frac{8}{2} \,\,and \,\,b^2=\frac{8}{8}$$
$$\therefore a=2\, \,and \,\,b=1$$
Hence area is $\pi\times 2\times 1=2\pi$ Square units.
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