What is the the area of the region enclosed by the equation

45x² + 20y² − 36xy = 2304

Solution: Area ~ 301.593

Using the form A x^2 + B xy + C y^2 = const, we note that
the conic discriminant, B^2-4AC, is negative, indicating 
an ellipse. With no D x and E y terms, the ellipse is centered at
the origin. 

N.b, the discriminant also equals (-const) here, 
which suggests there's a shortcut. However, not seeing
the shortcut, I go forward - giving details that help me. 

Using the Principle Axis Theorem to find the semi-major and 
minor axes, we can get the area. We express the LHS above by 
constructing matrix A, where {{x},{y}} is a column vector:

45 x^2 - 36 xy + 20 y^2 = {x,y} A {{x},{y}}

A = {{45,-18}, {-18,20}}, so

45 x^2 - 36 xy + 20 y^2
= {x,y}  {{45,-18}, {-18,20}}  {{x},{y}}

I believe we could have split the -36 into any two parts;
-18 was an arbitrary choice.

We seek eigenvectors X1, X2, and eigenvalues, l1,l2, s.t.

AX = lX
(AX-lI) X = 0 with I = {{1,0},{0,1}}

(Note: The matrix in () can not be invertible, or else we 
could do this: ()^-1 () X = 0 giving X=0, i.e. no eigenvectors. 
Since () is not invertible, det() = 0 )

det(AX-lI) = 0

det{{45-l,-18}, {-18,20-l}} = 0

l^2 - 65 l + 576 = 0

So our eigenvalues are: 

l_1,2 = 65/2 +/- sqrt(1921)/2 

Now solve for the eigenvectors:
AX = l1 A

(A - l1 I)X = 0

(A - l1 {{1,0},{0,1}})X = 0

row reduce the matrix:
{{45-(65/2+sqrt(1921)/2),-18},{-18,20-(65/2+sqrt(1921)/2)}}

= {{1,25/36 + sqrt(1921)/36},{0,0}}

So the first eigenvector is 

X1 = {1,25/36 + sqrt(1921)/36}

likewise, for l2,

row reduce the matrix: 
{{45-(65/2-sqrt(1921)/2),-18},{-18,20-(65/2-sqrt(1921)/2)}}

= {{1,25/36 - sqrt(1921)/36},{0,0}} 

X2 = {1,25/36 - sqrt(1921)/36}

These orthogonal vectors give the axes of the ellipse:
x+(1/36)(25 +/- sqrt(1921)y=0

witH the + for the major axis and the - for the minor axis

Rather than make them orthonormal, we just solve for 
their endpoint intersections with the ellipse: 

45 x^2 + 20 y^2 - 36 xy = 2304,

Doing the minor axis via substitution of x in terms of y. 

2304=(45/(36^2))[25+sqrt(1921)]^2 y^2 + 20 y^2 +
[25 + sqrt(1921) y^2

y = +/-3 sqrt{[2(9605 - 197 sqrt(1921)]/1921}
x = -(1/36) [25+ sqrt(1921) ] y

The minor axis terminates at +/- (-5.7660, 3.0158)

For the major axis

2304=(45/(36^2))[25-sqrt(1921)]^2 y^2 + 20 y^2 +
[25 - sqrt(1921) y^2

y = +/-3 sqrt{[2(9605 + 197 sqrt(1921)]/1921}
x = -(1/36) [25 - sqrt(1921) ] y

The major axis terminates at +/- (6.8377, 13.073)

The area is (pi a b), where a and b are the 
endpoints of the semi-minor and major axes.

Area = pi sqrt((-5.7660)^2+3.0158^2) sqrt(6.8377^2+13.073^2)

Area ~ 301.593

Edited on November 11, 2024, 9:33 pm

Posted by Steven Lord on 2024-11-10 18:53:26