199 Followers, Waiting for the another one.

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

So the area is $6\pi$ .

In the above I used the following:

1.The transformation equation for rotaion of $θ$ is $x=x'cosθ-y'sinθ$ and $y=x'sinθ+y'cosθ$ .

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

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

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

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

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

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

The area of the ellipse is $A = ab\pi = 2\cdot3\cdot \pi\ \dot{=}\ 18.84955592.$

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}