Find the area of the ellipse.

Question

Find the area of the ellipse $(a^2+1)x^2+2(a+b)xy+(b^2+1)y^2=c$ where $c>0$. I think the formula is $πab$. But when I tried to convert the 2nd degree equation to the standard form, replacing $x=(X \cos \theta - Y \sin \theta)$ and $y=(X \sin \theta + Y \cos \theta)$. And after vanishing $xy$ term I got $2 \theta = \tan^{-1} \frac{2}{a-b}$. And now I can't seem to figure out what to do.

Answer 1

Hint

Expand

$$(x+py)^2/r^2+y^2/s^2=1$$

$$x^2(s^2)+2ps^2xy+y^2(s^2+r^2)=r^2s^2$$

This will be identical to the given equation if

$$s^2/(a^2+1)=2ps^2/2(a+b)=(s^2+r^2)/(b^2+1)=r^2s^2/c$$

$$s^2=c/(a^2+1)$$

$$(s^2+r^2)/r^2s^2=(b^2+1)/c\implies r^2=?$$

Now the required area is $$\pi rs$$

Answer 2

The transformation from $(x,y)$ to $(X,Y)$ (depending on $\theta$) you have applied is a rotation of the coordinate axes. In order to avoid an $XY$-term you have chosen $\theta$ in the indicated way. In this way the new equation of the ellipse is of the "axis-aligned" form $$\Phi(a,b,\theta)X^2+\Psi(a,b,\theta)Y^2=c\ ,$$ where $\Phi$ and $\Psi$ are expressions in the indicated parameters. As $\theta$ depends in a known way on $a$ and $b$ you should be able to simplify this to $$\Phi_*(a,b)X^2+\Psi_*(a,b)Y^2=c\ .$$ You should be able to compute the area of this ellipse.