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Geometry Level 5

Find the area enclosed by the conic:

13 x 2 + 13 y 2 10 x y 60 2 x + 12 2 y + 72 = 0 \large{13x^2 + 13y^2 - 10xy -60 \sqrt{2} x + 12 \sqrt{2} y +72 = 0}

Details and Assumptions:

  • If you think that the conic is not a closed path, then type your answer as 0 0 .


The answer is 18.8495559215.

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3 solutions

Souryajit Roy
Sep 29, 2015

At first,rotate the the co-ordinate axes through 45 45 degrees(anticlockwise) to get ( x 3 ) 2 9 + ( y + 2 ) 2 4 = 1 \frac{(x'-3)^2}{9}+\frac{(y'+2)^2}{4}=1 .

So the area is 6 π 6\pi .

In the above I used the following:

1.The transformation equation for rotaion of θ θ is x = x c o s θ y s i n θ x=x'cosθ-y'sinθ and y = x s i n θ + y c o s θ y=x'sinθ+y'cosθ .

2.I chose θ θ to be 45 45 degrees to eliminate the x y xy term.

If the area exists, this must be an ellipse. First we center the conic at the origin: transform x = x u x' = x - u and y = y v y' = y - v so that the linear terms disappear.

This requires { 26 u 10 v = 60 2 , 10 u + 26 v = 12 2 ; \left\{\begin{array}{l} 26u - 10v = 60\sqrt2, \\ -10u + 26v = -12\sqrt 2\end{array}\right.; the solution is u = 2 1 2 2 , v = 1 2 2 u = 2\tfrac12\sqrt 2, v = \tfrac12\sqrt 2 .

In these new coordinates, the conic becomes 13 ( x ) 2 + 13 ( y ) 2 10 x y 72 = 0. 13(x')^2 + 13(y')^2 - 10 x'y' - 72 = 0.

To get rid of the mixed term x y x'y' , we rotate the coordinate system. Since the coefficients of x x' and y y' are equal, a rotation of 4 5 45^\circ will do. (More generally, one could find the eigenvectors of the coefficient matrix, ( 13 5 5 13 ) \left(\begin{array}{cc} 13 & -5 \\ -5 & 13\end{array}\right) , and use those as the new coordinate axes.)

Thus we set x = ξ + η 2 , y = ξ η 2 : x' = \frac{\xi + \eta}{\sqrt 2}, y' = \frac{\xi - \eta}{\sqrt 2}: 13 2 ( ξ + η ) 2 + 13 2 ( ξ η ) 2 10 2 ( ξ + η ) ( ξ η ) 72 \frac{13}{2}(\xi + \eta)^2 + \frac{13}{2}(\xi - \eta)^2 - \frac{10}{2}(\xi + \eta)(\xi - \eta) - 72 = 8 ξ 2 + 18 η 2 72 = 0. = 8\xi^2 + 18\eta^2 - 72 = 0. This is the equation of an ellipse, with semiaxes a = 72 8 = 3 , b = 72 18 = 2. a = \sqrt{\frac{72}{8}} = 3, b = \sqrt{\frac{72}{18}} = 2. (We chose the rotation ( x , y ) ( ξ , η ) (x',y') \mapsto (\xi,\eta) in such a way that lengths are preserved.)

The area of the ellipse is A = a b π = 2 3 π = ˙ 18.84955592. A = ab\pi = 2\cdot3\cdot \pi\ \dot{=}\ 18.84955592.

Lu Chee Ket
Oct 13, 2015

Rarely done but this is done. Not only a ratio of 3 : 2 but a = 3 and b = 2.

45 degrees turned from positive real axis before transition, centered at [5 Sqrt(2)/ 2, Sqrt(2)/ 2] after transition, this is an ellipse.

Transforms by rotation about the origin via parametric equations,

Cos Q and Sin Q are obtained in f(a, b, x, y) by Cramer's method and sum of their squares equals to 1 eliminates to form its equation. -10 x y is based to obtain ratio of a/ b = 3/ 2. (-26 p x + 10 q x) and (-26 q y + 10 p y) tells the center (p, q). [Turn, center and size.]

Constant term = 13 p^2 + 13 q^2 - 10 p q - 8 a^2 = 144 - 8 a^2 = 72 => a = 3.

To change the size of the transformed ellipse, just change the constant term of 72.

With a = 3 and b = 2,

Area enclosed = Pi (3)(2) = 6 Pi = 18.849555921538759430775860299677 {Square units}

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