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,

So, equation of ellipse now becomes,

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, $$8X^2+4Y^2=1$$

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?

  • 2
    $\begingroup$ i cant find any error , it seems right to me $\endgroup$
    – abkds
    Dec 16, 2013 at 16:40
  • $\begingroup$ The mistake is that $$\;\cos^2\theta-\sin^2\theta=\cos2\theta=0\iff 2\theta=\frac{(2n-1)\pi}2\iff\theta=\frac{(2n-1)\pi}4\;,\;\;n\in\Bbb Z$$ and not only $\;\theta=\frac\pi4\;$ ... $\endgroup$
    – DonAntonio
    Dec 16, 2013 at 16:56

2 Answers 2


You seem to have overseen that it can be $\;\theta=\frac{3\pi}4\;$ , and this indeed gives the desired result! :

$$\theta=\frac{3\pi}4\implies x=-\frac1{\sqrt2}(X+Y)\;\;,\;\;y=\frac1{\sqrt2}(X-Y)\implies\;\text{we get}$$

$$x^2+5y^2=1\implies a=1\;,\;\;b=\frac1{\sqrt5}$$

so the ellipse's area is

$$A=\frac\pi{\sqrt5}\implies \frac{3\sqrt5}\pi A=3$$


You need $3X^2+3Y^2$ instead of $6X^2+6Y^2$ in the middle of your argument.
It is $3X^2\cos^2\theta+3X^2\sin^2\theta$, not $3X^2+3X^2$


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