Enclosed Area 8-22-2020

For an ellipse or hyperbola of the form $Ax^2 +Bxy+Cy^2 +Dx+Ey+F=0$ , it can be transformed into the following standard form ( reference ):

$\frac {x'^2} {-S/(\lambda_1^2 \lambda_2)}+\frac {y'^2} {-S/(\lambda_1 \lambda_2^2)}=1$

where $S=\det\begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end {bmatrix}$ ; and $\lambda_1$ and $\lambda_2$ are eigenvalues of $\begin{bmatrix} A & B/2 \\ B/2 & C \end {bmatrix}$ or the roots of $\lambda^2 - (A+C) \lambda +(AC-(B/2)^2) =0$ . Putting in the values, we have

$S=\begin{vmatrix} 5 & 3/2 & - 1/2 \\ 3/2 & 1 & 2 \\ - 1/2 & -2 & - 3 \end {vmatrix} = - \frac {63}2$

$\begin{aligned} \lambda^2 - 6\lambda +\frac {11}4 & =0 \\ 4\lambda^2 - 24\lambda +11 & =0 \\ (2\lambda - 1) (2\lambda - 11) & =0 \\ \implies \lambda_1, \lambda_2 & = \frac 12, \frac {11} 2 \end{aligned}$

Since area of an ellipse is given by $\pi ab$ , where $a$ and $b$ are the semi-axes, then $\pi ab =-S(\lambda_1 \lambda_2)^\frac 32 \approx \boxed{21.7}$ .

One way to solve this numerically is to sweep $x$ over a range. For a particular value of $x$ , there is a quadratic equation to solve for $y$ . Call the two solutions $y_p$ and $y_n$ , where $y_p \geq y_n$ . The infinitesimal area is $dA = (y_p - y_n) dx$ . This method is applicable to any ellipse; even ones that do not contain the origin.

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import math

dx = 10.0**(-6.0)

x = -5.0

A = 0.0

flag = 0

while x <= 5.0:

    a = 1.0
    b = 3.0*x + 4.0
    c = 5.0*(x**2.0) - x - 3.0

    q = b**2.0 - 4.0*a*c

    if q >= 0.0:

        root = math.sqrt(q)

        yp = (-b + root)/(2.0*a)
        yn = (-b - root)/(2.0*a)

        if yp < yn:
            flag = 1

        dA = (yp-yn)*dx
        A = A + dA

        #print x,yp
        #print x,yn

    x = x + dx

print ""
print ""

print dx
print flag
print A

#1e-05
#0
#21.7000825478
#>>>

#1e-06
#0
#21.7000825386
#>>> 

We will make use of the following three facts :

Any transformation of axes system can be represented as the combination of translation and rotation .

It is immaterial which operation is executed first, translation or rotation. The final result is the same .

Transformation of axes system in a direction is equivalent to the same transformation of a figure in the opposite direction

Shifting the origin of coordinates from $(0,0)$ to $(\dfrac {14}{11},-\dfrac {43}{11})$

and then rotating the axes system counterclockwise through an angle of $\tan^{-1} (\frac 13)$ , the given equation of the curve is reduced to

$\dfrac {X^2}{\frac{252}{121}}+\dfrac {Y^2}{\frac{252}{11}}=1$

So, length of the major semi-axis is

$\sqrt {\dfrac {252}{121}}$

and the length of the minor semi-axis is

$\sqrt {\dfrac {252}{11}}$

Therefore the area enclosed by the curve is

$π\times \sqrt {\dfrac {252}{121}}\times \sqrt {\dfrac {252}{11}}$

$=\dfrac {252π}{11\sqrt {11}}\approx \boxed {21.7}$ .

Let $r = \begin{bmatrix} x && y \end{bmatrix}^T$ be a column vector representing the coordinates $x$ and $y$ , then the given equation can be written as

$r^T A r + b^T r + c = 0$

where,

$A = \begin{bmatrix} 5 && \frac{3}{2} \\ \frac{3}{2} && 1 \end{bmatrix}$

$b^T = \begin{bmatrix} -1 && 4 \end{bmatrix}$

$c = -3$

The first step is to eliminate the linear term ( $b^T r$ ), and to do that, we define

$r_0 = -\frac{1}{2} A^{-1} b$

Then, one can see that, similar to completing the square, we can write,

$r^T A r + b^T r = (r - r_0)^T A (r - r_0) - r_0^T A r_0$

To see why this is so, note that

$(r - r_0)^T A (r - r_0) - r_0^T A r_0 = r^T A r - 2 r^T A r_0 = r^T A r + r^T b = r^T A r + b^T r$

So now the given equation can be written as,

$(r - r_0)^T A (r - r_0) = -c + r_0^T A r_0 = d$

After calculating $r_0$ , the right hand side of the above equation comes to $d = 11 \frac{5}{11} = \frac{126}{11}$

To put the above in the standard format, we divide both sides by $d$ , and this gives,

$(r - r_0)^T ( \frac{1}{d} A ) (r - r_0) = 1$

One can check that matrix $A$ is positive definite (i.e. both its eigenvalues are positive), by noting that $A_{11} > 0$ and $| A | > 0$ , and therefore, the above equation represents an ellipse. We need to find the product of the lengths of the semi-axes, and this is equal to the square root of the reciprocal of the determinant of $\frac{1}{d} A$ . In other words,

$a b = \dfrac{1}{ \sqrt{ | \frac{1}{d} A | } }$

Since $A$ is $2 \times 2$ , then $| \dfrac{1}{d} A | = \dfrac{1}{d^2} | A |$

Hence,

$a b = \dfrac{d}{ \sqrt{ | A | } } = \dfrac{ \left( \dfrac{126}{11} \right) }{ \sqrt{ 5 - \frac{9}{4} } } = \dfrac{ \left( \dfrac{126}{11} \right) }{ \sqrt{ \frac{11}{4} } } = \dfrac{ 252 } { 11 \sqrt{11} }$

And the area is $A = \pi a b =\dfrac{ 252 \pi } { 11 \sqrt{11} } \approx 21.7$

From Wolfram Mathworld's Ellipse article :

$a=5;b=\frac{3}{2};c=1;d=-\frac{1}{2};e=2;f=-3;$

$a x^2+2 b x y+c y^2+2 d x+2 e y+f=0\to 5 x^2+3 x y-x+y^2+4 y-3=0$

$\Delta =\left| \begin{array}{ccc} a & b & d \\ b & c & e \\ d & e & f \\ \end{array} \right|\to -\frac{63}{2}$

$\Xi =\left| \begin{array}{cc} a & b \\ b & c \\ \end{array} \right|\to \frac{11}{4}$

$I=a+c\to 6$

$\frac{\Delta }{I}\to -(21/4)$

Since $\Delta \neq 0$ , $\Xi >0$ , and $\frac{\Delta }{I}<0$ , assuming the conic section is not degenerate (which if it were, then the problem would be nonsense), then it is an ellipse.

$\delta =b^2-a c\to -\frac{11}{4}$

The center of the ellipse is at $(\frac{\left| \begin{array}{cc} c & e \\ b & d \\ \end{array} \right| }{\delta }\to \frac{14}{11},\frac{\left| \begin{array}{cc} a & b \\ d & e \\ \end{array} \right| }{\delta }\to -\frac{43}{11})$

Translating the ellipse so that its center is at the origin, which does not change its area. This is done by substituting new values for the variables and simplifying.

$5 {x_1}^2+3 {x_1} {y_1}+{y_1}^2-\frac{126}{11}=0$

Note that linear terms are now gone.

The next problem is rotating the ellipse onto the axes, which also does not change the area.

The rotation angle is $\frac{1}{2} \tan ^{-1}\left(\frac{2 b}{a-c}\right)\to \left( \begin{array}{cc} \cos \left(\frac{1}{2} \tan ^{-1}\left(\frac{3}{4}\right)\right) & -\sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{3}{4}\right)\right) \\ \sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{3}{4}\right)\right) & \cos \left(\frac{1}{2} \tan ^{-1}\left(\frac{3}{4}\right)\right) \\ \end{array} \right)$ , in which you can see the Pythagoras 3-4-5 triangle, which permits that rotation matrix to be simplified in turn to $\left( \begin{array}{cc} \frac{3}{\sqrt{10}} & -\frac{1}{\sqrt{10}} \\ \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \end{array} \right)$ .

After substituting the rotated coordinates and simplifying: $\frac{1}{11} \left(11 \left(11 {x_2}^2+{y_2}^2\right)=252\right)$

By substituting zero for x2 and y2 in turn, the semi-radii are $\frac{6 \sqrt{7}}{11}$ and $6 \sqrt{\frac{7}{11}}$ , respectively.

Therefore, the area is $\pi 6 \frac{\sqrt{7}}{11} 6 \sqrt{\frac{7}{11}}\to \frac{252 \pi }{11 \sqrt{11}}\approx 21.7001$