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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.

Tips
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  • First of all $2x^2+6xy+5y^2$ is not an equation. That aside, apply a rotation with unknown sine and cosine to this and use the fact that the sum of their squares is equal to one to find the correct values to eliminate the cross term. If the resulting expressions are complicated, you’ll just have to deal with them. A simpler approach is to compute the eigenvalues of this quadratic form, but those expressions will also involve radicals. – amd Jan 11 '19 at 20:56
  • Ok I try to solve the resulting expression. – Tips Jan 12 '19 at 03:28

2 Answers2

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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 $x^2+y^2 = \dfrac{7+3 \sqrt 5}{2}$

Hari Shankar
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To avoid rotations, you could find the intersection $P$ between the ellipse (which is centred at $O=(0,0)$) and a generic line through the centre $y=mx$. Length $PO$ is then an expression in $m$ whose maximum gives the radius of the auxiliary circle.

Intelligenti pauca
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