Common area of two ellipses

Solving the two ellipse equations, the points of intersection are $P(1, 4)$ , $Q(5, 2)$ , $R(3, -2)$ , and $S(-1, 0)$ .

Using the distance formula, the sides of quadrilateral $PQRS$ are $PQ = QR = RS = PR = \sqrt{20}$ and the diagonals are $PR = QS = \sqrt{40}$ , therefore $PQRS$ is a square with an area of $A_{PQRS} = 20$ .

The equation of the line with $PQ$ is $y = -\frac{1}{2}x + \frac{9}{2}$ and the equation of the line with $PR$ is $y = 2x + 2$ .

The equation of the upper part of the blue ellipse rearranges to $y = \frac{1}{13}(5\sqrt{-3x^2 + 12x + 40} - 4x + 21)$ , and the equation of the upper part of the orange ellipse rearranges to $y = \frac{1}{6}(5\sqrt{-2x^2 + 8x + 19} + 7x - 8)$ .

The area of the section between the blue ellipse and $PQ$ is then $A_{PQ} = \int_{1}^{5} ((\frac{1}{13}(5\sqrt{-3x^2 + 12x + 40} - 4x + 21)) - (-\frac{1}{2}x + \frac{9}{2})) dx$ , and the area of the section between the orange ellipse and $PR$ is $A_{PR} = \int_{-1}^{1} ((\frac{1}{6}(5\sqrt{-2x^2 + 8x + 19} + 7x - 8)) - (2x + 2)) dx$ .

By symmetry, $A_{RS} = A_{PQ}$ and $A_{QR} = A_{PR}$ , so the total shaded area is $A = 2A_{PQ} + 2A_{PR} + A_{PQRS} \approx 22.9055$ , which makes $\lfloor 1000A \rfloor = \boxed{22905}$ .

I will take advantage of the graph provided to get a much simpler solution.

Rotate the ellipses with respect to their center $(2,1)$ so that the coordinates of the intersection points are $(\pm \sqrt{5}, \pm \sqrt{5})$ . In these rotated axes centered at $(2,1)$ the ellipses have equations of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where the major semi-axis $a$ of each one is read from the graph and $b$ can be calculated substituting one of the intersection points. After little manipulation, the ellipses become:

• Blue ellipse: $x^2+3y^2=20$ , with major semi-axis $a=2\sqrt{5}$
• Orange ellipse: $y^2+8x^2=45$ , with major semi-axis $a=3\sqrt{5}$

The square has an area of $20$ . We just need to calculate the extra area limited by the sides of the square and the four arches.

We get integrals of the form $\displaystyle \int_{-x_0}^{x_0}\sqrt{a^2-x^2}\, dx$ which are quite straightforward to solve using the variable change $x=a\sin\theta, dx=a\cos\theta \, d\theta$ :

$\displaystyle \int_{-x_0}^{x_0}\sqrt{a^2-x^2}\, dx=2 \times \displaystyle \int_{0}^{\arcsin\left(\frac{x_0}{a}\right)}\cos^2\theta \, d\theta=2 \times \displaystyle \int_{0}^{\arcsin\left(\frac{x_0}{a}\right)}\cos^2\theta \, d\theta=2 \times \displaystyle \int_{0}^{\arcsin\left(\frac{x_0}{a}\right)}\dfrac{1}{2}+\dfrac{1}{2}\cos2\theta \, d\theta=\left[ \theta+\dfrac{1}{2}\sin2\theta \right]_0^{\arcsin\left(\frac{x_0}{a}\right)}$

Using $x_0=\sqrt{5}$ and the values of $a$ for each ellipse we obtain:

• $A_{blue}\left[ \theta+\dfrac{1}{2}\sin2\theta \right]_0^{\arcsin\left(\frac{\sqrt{5}}{2\sqrt{5}}\right)}\approx 1.045998$
• $A_{orange}\left[ \theta+\dfrac{1}{2}\sin2\theta \right]_0^{\arcsin\left(\frac{\sqrt{5}}{3\sqrt{5}}\right)}\approx 0.406772$

The total area is $A=A_{square}+2 \times A_{blue}+2 \times A_{orange}\approx 22.905540 \implies \boxed{\lfloor 1000A \rfloor=22905}$

The given equation of the first ellipse can be written using matrices as

$r^T A r + b^T r + c = 0 \hspace{24pt} (1)$

where

$r = \begin{bmatrix} x && y \end{bmatrix}^T$

$A = \begin{bmatrix} 7 && 4 \\ 4 && 13 \end{bmatrix}$

$b^T =\begin{bmatrix} -36 && -42 \end{bmatrix}$

$c = -43$

The first step is to find the center of this ellipse. It is given by

$r_0 = -\frac{1}{2} A^{-1} b \hspace{24pt} (2)$

One can verify that equation (1) be re-written as,

$( r - r_0 )^T A (r - r_0) = - c + r_0^T A r_0 \hspace{24pt} (3)$

Define the matrix $Q$ as follows

$Q = \dfrac{ A }{- c + r_0^T A r_0} \hspace{24pt} (4)$

then, from equation (3) and (4), we obtain

$(r - r_0)^T Q (r - r_0) = 1 \hspace{24pt} (5)$

The next step is to diagonalize $Q$ , and write it as $Q = R D R^T$

The procedure is as follows,

Let $\hspace{6pt} \theta = \dfrac{1}{2} \tan^{-1} \dfrac{ 2 Q_{12} }{Q_{11}-Q_{22}} \hspace{24pt} (6)$

Then

$R = \begin{bmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} \hspace{24pt} (7)$

and

$D = \text{diag} \{ d_{11} , d_{22} \} \hspace{24pt} (8.1)$

with

$d_{11} = Q_{11} \cos^2 \theta + Q_{22} \sin^2 \theta + 2 Q_{12} \cos \theta \sin \theta \hspace{24pt} (8.2)$

$d_{22} = Q_{11} \sin^2 \theta + Q_{22} \cos^2 \theta - 2 Q_{12} \cos \theta \sin \theta \hspace{24pt} (8.3)$

Replacing matrix $Q$ in equation (5) by its diagonalized form, we obtain

$(r - r_0)^T R D R^T (r - r_0) = 1 \hspace{24pt} (9)$

For an ellipse, the diagonal matrix $D$ has positive entries, and therefore can be factored as $D = D^{\frac{1}{2}} D^{\frac{1}{2}}$

Using this factorization, and defining the vector

$u = D^{\frac{1}{2}} R^T (r - r_0) \hspace{24pt} (10)$

then it follows that $u^T u = 1$ , thus vector $u$ is a unit vector, and can therefore be parameterized as $u = [\cos t , \sin t]^T$ .

Solving for $r$ in terms of $u$ gives,

$r = r_0 + R D^{-\frac{1}{2}} u = r_0 + V u \hspace{24pt} (11)$

The columns of matrix $V$ are vectors extending along the major and minor axes directions and having lengths of the semi-major

and semi-minor axes.

The same procedure can be applied to the second ellipse, and we'll end up with the following parametric equations for

the two ellipses,

$r_1(t) = r_{C1} + V_1 u_1(t) \hspace{24pt} (12.1)$

$r_2(s) = r_{C2} + V_2 u_2(s)\hspace{24pt} (12.2)$

Performing the calculations detailed above, the reader can verify that,

$r_{C1} = r_{C2} = \begin{bmatrix} 2 && 1 \end{bmatrix}^T$

$V_1 = \begin{bmatrix} 4 && \dfrac{2}{\sqrt{3}} \\ -2 && \dfrac{4}{\sqrt{3}} \end{bmatrix}$

$V_2 = \begin{bmatrix} \dfrac{3}{\sqrt{2}} && 3 \\-\dfrac{3}{2\sqrt{2}} && 6 \end{bmatrix}$

We next have to find the points of intersection of the two ellipses, so we set up the equation,

$r_1(t) = r_2(s) \Longrightarrow r_{C1} + V_1 u_1(t) = r_{C2} + V_2 u_2(s) \hspace{24pt} (13)$

Solving for $u_1(t)$ , we obtain,

$u_1(t) = V_1^{-1} ( r_{C2} - r_{C1} + V_2 u_2(s) ) = e + F u_2(s) \hspace{24pt} (14)$

where $e = V_1^{-1} (r_{C2} - r_{C1})$ and $F = V_1^{-1} V_2$

Since $u_1$ is a unit vector, we have $u_1^T u_1 = 1$ , and this leads to,

$(e + F u_2 )^T (e + F u_2 ) = 1 \hspace{24pt} (15)$

Expanding the left hand side of (15),

$u_2^T F^T F u_2 + 2 u_2^T F^T e + e^T e = 1 \hspace{24pt} (16)$

For reference, let's define $G = F^T F , h = 2 F^T e, q = e^T e - 1$ , then equation (16) can be written as,

$u_2^T G u_2 + u_2^T h + q = 0 \hspace{24pt} (17)$

Recall that $u_2 = [ \cos s, \sin s ]^T$ , thus expanding equation (17) yields,

$G_{11} \cos^2 s + G_{22} \sin^2 s + 2 G_{12} \sin s \cos s + h_1 \cos s + h_2 \sin s + q = 0 \hspace{24pt} (18)$

Using the identities $\cos^2 s = \frac{1}{2} (1 + \cos(2s) )$ , and $\sin^2 s = \frac{1}{2} (1 - \cos(2s) )$ as well as $2 \sin s \cos s = \sin(2s)$ , equation (18) becomes,

$c_1 \cos s + c_2 \sin s + c_3 \cos 2 s + c_4 \sin 2 s + c_5 = 0\hspace{24pt} (19)$

where $c_1 = h_1, c_2 = h_2 , c_3 = \frac{1}{2} ( G_{11} - G_{22} ) , c_4 = G_{12} , c_5 = q + \frac{1}{2} (G_{11}+G_{22})$

After performing all the calculations above, equation (19) produces four solutions, and they are,

$s_1 = 0.339836909 \hspace{12pt}s_2=2.801755744\hspace{12pt} s_3=3.481429563\hspace{12pt} s_4=5.943348398$

The corresponding values for parameter $t$ of the first ellipse are,

$t_1=1.047197551 \hspace{12pt}t_2=2.094395102\hspace{12pt}t_3=4.188790205\hspace{12pt}t_4=5.235987756$

The four solution for $s$ (and also the four corresponding solutions of $t$ ) correspond in turn to the Cartesian points:

$P_1 = (5, 2) ,\hspace{12pt} P_2 = (1, 4),\hspace{12pt} P_3 = (-1, 0),\hspace{12pt} P_4 = (3, -2)$

The region common to both ellipses is bounded by the first ellipse for $t \in [t_1 , t_2]$ and for $t \in [t_3, t_4]$ and is bounded by the second ellipse for $s \in [s_2, s_3]$ and for $s \in [s_4, s_1 + 2\pi ]$

The area enclosed by the parametric curve $r(t) = (x(t), y(t) )$ whose $t$ -derivative is $r'(t) = (x'(t), y'(t) )$ is given by the formula

$\text{Area} = \dfrac{1}{2} \left| \displaystyle \int_{t_i}^{t_f} x(t) y'(t) - x'(t) y(t) dt \right| \hspace{24pt}(20)$

If we extend the vector $r(t)$ and its $t$ -derivative into three dimensions, by padding our $r(t)$ and $r'(t)$ with a zero $z$ coordinate, then the above integral can be written conveniently using cross product notation as,

$\text{Area} = \dfrac{1}{2} \left| \displaystyle \int_{t_i}^{t_f} \left( r(t) \times r'(t) \right)_z dt \right| \hspace{24pt}(21)$

where $\left( \cdot \right)_z$ is the $z$ coordinate of its vector argument. We have to use this formula four times corresponding to the four intervals that we have. Now let's see what this evaluation entails. We have

$r(t) = r_{C} + V u$

If $V = [ v_1 \hspace{6pt} v_2 ]$ , where $v_1$ and $v_2$ are the columns of $V$ , then we have,

$r(t) = r_{C} + v_1 \cos t + v_2 \sin t$

and its $t$ -derivative is,

$r'(t) = - v_1 \sin t + v_2 \cos t$

Hence,

$r(t) \times r'(t) = - (r_{C} \times v_1) \sin t + (r_{C} \times v_2) \cos t + ( v_1 \times v_2 ) ( \cos^2 t + \sin^2 t)$

which simplifies to,

$r(t) \times r'(t)=- (r_{C} \times v_1) \sin t + (r_{C} \times v_2) \cos t + v_1 \times v_2$

Evaluating the area integral over a given interval $[t_i , t_f ]$ yields,

$A_k = \dfrac{1}{2} \left( (r_{C} \times v_1)_z (\cos t_f - \cos t_i) + (r_{C} \times v_2)_z (\sin t_f - \sin t_i) + ( v_1 \times v_2 )_z (t_f - t_i) \right) \hspace{10pt} k = 1, 2, 3, 4$

In this particular problem, and since $r_{C1} = r_{C2}$ we can shift the center of our parametric curves to the origin, and this simplifies the above formula into

$A_k =\dfrac{1}{2} (v_1 \times v_2 )_z (t_f - t_i ) \hspace{24pt} k = 1, 2, 3, 4$

From matrix $V_1$ and $V_2$ above, we have, for the first ellipse, $( v_1 \times v_2 )_z = \dfrac{20}{\sqrt{3}}$ , and for the second ellipse, $(v_1 \times v_2)_z = \dfrac{45}{2\sqrt{2}}$

Plugging in the $t$ and $s$ values we found, the area is given by,

$\text{Area} = \dfrac{1}{2} \left( \dfrac{20}{\sqrt{3}} ( t_2 - t_1 + t_4 - t_3 ) + \dfrac{45}{2\sqrt{2}} ( s_3 - s_2 + s_1 + 2 \pi - s_4 )\right)$

This evaluates to, $\text{Area} = A = 22.9055400043$ , and this makes the answer $\lfloor 1000A \rfloor = \boxed{22905}$ .