Question:
If $B^2-4AC \lt 0$, the equation
$$Ax^2+Bxy+Cy^2=1$$
represents an ellipse. Prove that the area of the ellipse is
$\frac{2\pi}{\sqrt{4AC-B^2}}$.
I know that, if $2a$ and $2b$ are the major and minor axis of the ellipse respectively, its area is $\pi ab$.
I honestly have no idea how to go about this. Any help is appreciated.
Best Answer
Hint For a linear transformation $T : \Bbb R^2 \to \Bbb R^2$ given by $$\pmatrix{u\\v} = \pmatrix{a&b\\c&d} \pmatrix{x\\y} ,$$ the areas of a (nice) region $E \subset \Bbb R^2$ (e.g., our ellipse) and its image $T(E) \subset \Bbb R^2$ are related by $$\textrm{area}(T(E)) = |{\det T}| \,\textrm{area}(E) .$$ On the other hand, we know that there is a linear transformation $T$ that maps our ellipse to the unit ball, and so for such a transformation rearranging the previous equation gives $$\textrm{area}(E) = \frac{\textrm{area}(T(E))}{|{\det T}|} = \frac{\pi}{|{\det T}|} .$$ So, we need only write $|{\det T}|$ in terms of $A, B, C$, and in particular show that $$|{\det T}| = |a d - b c| = \frac{1}{2} \sqrt{4 A C - B^2} .$$