Area of ellipse which is not in standard form

Question

By graphing device i understand that $x^2+xy+y^2=1$ is ellipse. By some geometry i find area of above ellipse which comes out $\pi$ (is it right?), but it was easy case. Is there any quick method or standard formula to calculate it or we have to convert it into standard form always and then calculate?

Answer 1

We know that such an ellipse must be symmetric about the line $x = y$, since interchanging the variables gives us the same equation. Thus one axis is along this line, and solving the system $$\begin{align*} x^2 + xy + y^2 &= 1, \\ x &= y, \end{align*}$$ gives $$(x,y) = \pm (1/\sqrt{3}, 1/\sqrt{3}).$$ This means the length of this axis is $2 \sqrt{2/3}$. We can also verify in a similar fashion that the ellipse is invariant with respect to the transformation $(x,y) \to (-y, -x)$; that is to say, the perpendicular axis is found by solving the system $$\begin{align*} x^2 + xy + y^2 &= 1, \\ x &= -y, \end{align*}$$ which gives $$(x,y) = \{(-1,1), (1,-1)\}.$$ Thus this axis is the major axis and has length $2\sqrt{2}$; the ellipse has total area $$\pi ab = \pi (\sqrt{2})( \sqrt{2/3}) = 2\pi/\sqrt{3}.$$

Of course, this was a method that is unique to this case only. But I wanted to show that an elementary approach can be a useful check against the general approach of coordinate transformations.

Answer 2

Let us assume to have the conic $Ax^2+2Bxy+Cy^2=1$ associated with the symmetric matrix $$ M=\begin{pmatrix}A & B \\ B & C\end{pmatrix}. $$ Such conic is an ellipse iff $A>0, C>0$ and $\det M=AC-B^2>0$. $\det M$ equals the product $\lambda_1 \lambda_2$ of the (real, positive) eigenvalues of $M$ and by the spectral theorem the area enclosed by the ellipse is given by $$ \iint_{(x\, y)M(x\, y)^T \leq 1} 1\,dx\,dy = \iint_{\lambda_1 x^2+\lambda_2 y^2\leq 1}1\,dx\,dy =\frac{1}{\sqrt{\lambda_1\lambda_2}}\iint_{x^2+y^2\leq 1}1\,dx\,dy.$$ It follows that under the previous constraints, the area enclosed by $Ax^2+2Bxy+Cy^2=1$ is given by $$\mathcal{A}=\color{red}{\frac{\pi}{\sqrt{AC-B^2}}}=\frac{2\pi}{\sqrt{-\Delta}}.$$ It is straightforward to check that $A=C=1$ and $B=\frac{1}{2}$ fulfill the above constraints, hence the area enclosed by the ellipse $x^2+xy+y^2=1$ is given by $\color{red}{\frac{2\pi}{\sqrt{3}}}$.

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Remark: you do not need to compute the spectral form of $M$ to find the area enclosed by your ellipse, $\det M$ is enough.

Answer 3

The ellipse after the rotation of $45°$ has an equation

$\dfrac{X^2}{2}+\dfrac{Y^2}{\frac{2}{3}}=1$

Thus the area is $A=\sqrt{2}\sqrt{\dfrac{2}{3}}\pi=\dfrac{2}{\sqrt 3}\pi\approx 3.6276$

Answer 4

If the conic $ Q(x,y) = ax^2 + by^2 + c + 2hxy + 2fy + 2gx$ is an ellipse, then if we translate the orgin of our coordinate system to the centre of $ Q$ to get,

$$ {R(X,Y) = Q(X + u, Y + v)} = aX^2 + b Y^2 + 2hXY + c^\prime$$

where $(u, v)$ is the centre of the conic given. We need to find $(u, v)$ and $c^\prime$ (practically we only need $c'$).

Anyway $(u, v)$ is the solution of $$\begin{cases}\dfrac{\partial Q}{\partial x} = 0 \\ \dfrac{\partial Q}{\partial y} = 0 \end{cases}$$

and $c^\prime$ is given by $ c^\prime =\dfrac{\Delta}{\delta}$ where $$\delta = \begin{vmatrix} a & h \\ h & b\end{vmatrix}.$$ and $$\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}.$$

The problem reduces to finding the area of the curve $R(X, Y) = 0$, which is equivalent to finding the area of $ S (X, Y) = \dfrac{-a}{c^\prime}X^2 + \dfrac{-b}{c^\prime} Y^2 + 2\dfrac{-h}{c^\prime}XY - 1$.

Let $\alpha = \dfrac{-a}{c^\prime} , \beta = \dfrac{-b}{c^\prime}$ and $\gamma = \dfrac{-h}{c^\prime}$, then we have $ S(X, Y) = \alpha X^2 + \beta Y^2 + 2\gamma XY - 1$.

To write $ S$ in standard form we need to eliminate $ XY$ term. If $ \gamma = h \ne 0$ then that means the ellipse is rotated by some angle $\theta$ given by the relation, $$\tan (2\theta) = \dfrac{2\gamma}{ \alpha - \beta} = \dfrac{2h}{a - b}.$$

So to remove $XY$ term we rotate $S$ by $-\theta$, $${X \choose Y} = \left[\begin{matrix}\cos\theta & -\sin \theta \\ \sin \theta& \cos \theta \end{matrix}\right] {X^\prime \choose Y^\prime}.$$

We will then get,

$$ S(X^\prime, Y^\prime) = \dfrac{X{{^\prime}^2} }{\alpha'^2} + \dfrac{Y{{^\prime}^2} }{\beta'^2} - 1 = 0$$

The area of which is given by $\pi \alpha' \beta'$.

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For example, Let $Q(x,y) = x^2 + 2y^2 + 2xy + 4x + 4y + 1$,

Here we have $c^\prime = -3$ and $\theta = \dfrac12 \tan^{-1} (-2)$.

First translating the origin we get, $S(X, Y) = \dfrac{1}{3} X^2 + \dfrac23 Y^2 + 2\dfrac13XY - 1$.

Now we rotate the cooridinate system by $\dfrac12 \tan^{-1} (2)$ to get the equation of the ellipse as,

$$\dfrac{X'^2}{\dfrac6{3- \sqrt{5}}} + \dfrac{Y'^2}{\dfrac6{3+ \sqrt{5}}} = 1 $$

So the area is $3\pi$.

Not the easiest method but the simplest method for sure.

Answer 5

Let $x' = x + \frac12y.$ Then $ x'^2 = x^2 + xy + \frac14y^2, $ so $$ x'^2 + \frac34 y^2 = x^2 + xy + y^2 = 1. $$

The transformation $T: (x,y) \mapsto (x',y)$ is a skew transformation; for example, it maps the square $(0,0), (1,0), (1,1), (0,1)$ to the parallelogram $(0,0), (1,0), (\frac32,1), (\frac12,1).$ The transformation $T$ does not change the areas of the regions it acts on.

Now if $E$ is the ellipse that satisfies $ x^2 + xy + y^2 = 1,$ then $TE$ (the image of $E$ under the transformation $T$) satisfies the formula $$ x^2 + \frac34 y^2 = 1. $$ Since $T$ preserves area, to find the area of $E$ we need only find the area of $TE.$ But as we can see, $TE$ is an ellipse with semi-major axes $1$ and $\frac{2}{\sqrt3},$ whose area we can easily compute.