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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

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  • $\begingroup$ $4\sin\theta\cos\theta=2\sin2\theta.$ $\endgroup$ – Jyrki Lahtonen Oct 11 '14 at 13:00
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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

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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.

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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!

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  • $\begingroup$ The reason why polar coordinates work ($\Leftrightarrow$ that $2\theta$ appears) is that this time the axes are tilted by 45 degrees. Everything fits as $\sin2\theta$ attains it maximum at $\theta=\pi/4$. $\endgroup$ – Jyrki Lahtonen Oct 12 '14 at 13:45

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