<s>[INST] Equation of auxiliary circle of the ellipse $2x^2 +6xy + 5y^2$ =1 Equation of auxiliary circle of the ellipse $2x^2 +6xy + 5y^2$= 1
My approach is , First I try remove xy term from the equation, to convert the given equation in the standard equation of ellipse and find the value of $a$ and $b$. For this , I use the concept of rotation of axis, using this concept first I find $\tan2\theta = -2$ then I find the value of $\sin \theta$ and $\cos \theta$. But the problem is the the value of $\sin \theta$ and $\cos\theta$ are complex and when I try to solve using the concept of rotation of axis but equation become too much complex and also take a lots  of time to solve it. So I please tell me another approach or method to solve this question. 
 [/INST] The auxiliary circle is centered at the center of the ellipse and its radius equals the semi-major axis of the ellipse. Any conic of the form $ax^2+2hxy+by^2=1$ is centered at origin (as if $(x,y)$ lies on the ellipse so does $(-x,-y)$).
To find the semi-major axis we need to find the greatest distance of any point on the conic from the center. So we turn to polar coordinates $x=r \cos \theta, y= r\sin \theta$
Substituting in the equation we get $r^2 = \dfrac{1}{2 \cos^2 \theta+6 \sin \theta \cos \theta+ 5 \sin^2 \theta} = \dfrac{2}{7+6\sin 2 \theta - 3 \cos 2 \theta} \le \dfrac{7+3 \sqrt 5}{2}$
Hence the equation of the auxiliary circle is $x^2+y^2 = \dfrac{7+3 \sqrt 5}{2}$
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