Prove that area of ellipse $Ax^2+Bxy+Cy^2=1$ is equal to $\frac{2\pi}{\sqrt{4AC-B^2}}$

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 $$\pi ab$$.

• Solve the equation for $y$ in terms of $x$. Integrate the upper value minus the lower value, over the interval where the solutions are real. Nov 1 '17 at 13:38
• How do i get rid of the shared value $Bxy$ ?? Nov 1 '17 at 13:44
• Solve for y in terms of $x$ ... quadratic formula ... $$Cy^2 + (Bx)y + (Ax^2-1) = 0 \\ y = \frac{-Bx\pm\sqrt{4C-(4AC-B^2)x^2}}{2C}$$ Nov 1 '17 at 15:22
• Ohhh okay thanks, I'll try that! Nov 1 '17 at 18:41
• For future questions, please provide more background information. When there are multiple ways to attack a problem, like here, it’s hard to give you an answer you’ll understand without knowing what you’ve got to work with.
– amd
Nov 1 '17 at 19:39

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 $$|{\det T}| = |a d - b c| = \frac{1}{2} \sqrt{4 A C - B^2} .$$

Additional hint Substituting using the transformation formula for $$T$$ gives that in $$xy$$-coordinates the unit circle is $$(a^2 + c^2) x^2 + 2(ab + cd) x y + (b^2 + d^2) y^2 .$$ Thus, the desired transformation satisfies $$A = a^2 + c^2, \quad B = 2 (a b + c d), \quad C = b^2 + d^2.$$

• Sorry, I dont really understand this. I dont think I have studied the materials necessary to understand it. But thanks anyways, maybe someone else will instead :) Nov 1 '17 at 18:20
• What course are you taking? If it's the integral symbols (calculus) that you find daunting, they're not really necessary: It's enough to know that, given a linear transformation $T$, the areas of the regions $E$ and $T(E)$ are related by $\textrm{area}(T(E)) = |{\det T}| \textrm{area}(E)$. (In fact, perhaps I'll edit the answer to reflect this.) By design, in our case $T(E)$ is a unit circle, so $\textrm{T(E)} = \pi$, which is enough to give the second display equation. Nov 1 '17 at 21:05
• I don't have issues with integral calculus , rather I've never really taken Linear transformation and how they relate to areas ( atleast I don't think I've taken linear transformation) and matrices so I dont know how to use those. I'm a senior high-school student Nov 1 '17 at 21:10
• What course is this from them? Do you have a particular related results available? Nov 1 '17 at 21:16
• i dont have any related results, and Where I come from ( a region in the middle east) we don't have "courses" per say, rather we just study the book as is, so I doubt I can find any info on that. edit : actually, I just checked last year's book and it did have matrices , but looks like we just skipped the subject like many of the other ones due to alot of time restraints (schools started pretty late last year due to many issues) Nov 1 '17 at 21:29

In polar coordinates, the ellipse is given by $$r^2 (\theta)= \frac1{A\cos^2\theta +B\cos\theta\sin\theta +C\sin^2\theta}$$

Denote $$\Delta = \sqrt{(A-C)^2+B^2}$$ and integrate its area as follows

\begin{align} A=\frac12 \int_0^{2\pi} r^2(\theta)\> d\theta &= \frac12 \int_0^{2\pi} \frac{1}{A\cos^2\theta +B\cos\theta\sin\theta +C\sin^2\theta}\> d\theta\\ &= \int_0^{2\pi} \frac{1}{(A+C)+(A-C)\cos 2\theta +B\sin2\theta}\> d\theta\\ &= \int_0^{2\pi} \frac{1}{(A+C)+\Delta\cos 2\theta }\>d\theta\\ &= 4\int_0^{\pi/2} \frac{d(\tan\theta)}{(A+C+\Delta)+(A+C-\Delta)\tan^2\theta }\\ &=\frac{2\pi}{\sqrt{(A+C)^2 - \Delta^2}} =\frac{2\pi}{\sqrt{4AC-B^2}}\\ \end{align}