If $B^2-4AC \lt 0$, the equation

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.

  • $\begingroup$ 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. $\endgroup$
    – GEdgar
    Nov 1 '17 at 13:38
  • $\begingroup$ How do i get rid of the shared value $Bxy$ ?? $\endgroup$
    – Dahen
    Nov 1 '17 at 13:44
  • $\begingroup$ Solve for y in terms of $x$ ... quadratic formula ... $$ Cy^2 + (Bx)y + (Ax^2-1) = 0 \\ y = \frac{-Bx\pm\sqrt{4C-(4AC-B^2)x^2}}{2C} $$ $\endgroup$
    – GEdgar
    Nov 1 '17 at 15:22
  • $\begingroup$ Ohhh okay thanks, I'll try that! $\endgroup$
    – Dahen
    Nov 1 '17 at 18:41
  • 1
    $\begingroup$ 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. $\endgroup$
    – 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 $$A = a^2 + c^2, \quad B = 2 (a b + c d), \quad C = b^2 + d^2.$$

  • $\begingroup$ 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 :) $\endgroup$
    – Dahen
    Nov 1 '17 at 18:20
  • $\begingroup$ 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. $\endgroup$ Nov 1 '17 at 21:05
  • $\begingroup$ 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 $\endgroup$
    – Dahen
    Nov 1 '17 at 21:10
  • $\begingroup$ What course is this from them? Do you have a particular related results available? $\endgroup$ Nov 1 '17 at 21:16
  • $\begingroup$ 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) $\endgroup$
    – Dahen
    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}


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