Show that the equation $(a^2+1)x^2 + 2(a+b) xy + (b^2+1)y^2 =c$ represents an ellipse

Question

Show that the equation:

$$(a^2+1)x^2 + 2(a+b) xy + (b^2+1)y^2 =c$$

in which $c>0$ represents an ellipse of area $\dfrac{\pi c}{ab-1}$.

I attempt to show $(a^2+1)(b^2+1)-(a+b)^2>0$ which means $\Delta<0$.But how could I do this without knowing any information of $a$ and $b$?

Please anyone help me to show that the equation represents an ellipse of area $\dfrac{\pi c}{ab-1}$.

Thanks in advance.

Answer 1

A quicker way is to re-arrange the conic:

\begin{align} c &= (ax+y)^2+(x+by)^2 \\ c &= u^2+v^2 \\ u &= ax+y \\ v &= x+by \\ du \, dv &= \frac{\partial (u,v)}{\partial (x,y)} \, dx \, dy \\ \pi c &= A \begin{Vmatrix} a & 1 \\ 1 & b \end{Vmatrix} \\ A &= \frac{\pi c}{|ab-1|} \end{align}

• $u$ and $v$ are two skew axes re-scaled respectively by $\sqrt{a^2+1}$ and $\sqrt{b^2+1}$ and made by an angle of $\theta=\sin^{-1} \dfrac{|ab-1|}{\sqrt{(a^2+1)(b^2+1)}}$.

• For $ab=1$, the conic degenerates into two parallel lines.

Answer 2

Very good approach, here is a proof that for almost all $a,b$ the innequality is true;$$(a^2+1)(b^2+1)-(a+b)^2=a^2b^2+1-2ab=(ab-1)^2\geq0$$The only way that they are equal, is if $ab=1\to a=\frac1b$. In this case it's not representing a elipse.