Area sum

In response to the Challenge Master's note:

According to Discriminant of a Conic Section , the general equation of a conic section is given below if $\Delta \ne 0$ .

$f(x,y)=ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$

$\Delta =\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}= abc+2fgh-a{ f }^{ 2 }-b{ g }^{ 2 }-c{ h }^{ 2 }$

If the discriminant $h^2 - ab < 0$ and $h \ne 0$ , then the conic section is an ellipse.

Now, we have: $f(x) = 5x^2 + 6xy + 5y^2 - 32 = 0$ .

$\Delta = =\begin{vmatrix} 5 & 3 & 0 \\ 3 & 5 & 0 \\ 0 & 0 & -32 \end{vmatrix} = -512 \ne 0$

The discriminant: $h-ab = 3-25 < 0$ and $h \ne 0$ , therefore, $f(x)$ is an ellipse.

For an ellipse of the form $\alpha x^2 + \beta xy + \gamma y^2 = 1$ , then the area of the ellipse (eqn. 88) is given by $A = \dfrac {2\pi}{\sqrt{4\alpha \gamma - \beta^2}}$ .

$\begin{aligned} 5x^2 + 6xy + 5y^2 & = 32 \\ \frac 5{32}x^2 + \frac 6{32} xy + \frac 5{32} y^2 & = 1 \end{aligned}$

$\begin{aligned} \implies A & = \frac {2\pi}{\sqrt{4\cdot \frac {5^2}{32^2} - \frac {6^2}{32^2}}} \\ & = 8 \pi \approx \boxed{25.133} \end{aligned}$

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Previous solution

The curve is an ellipse with semi-major and minor axes tilted $-45^\circ$ with x- and y-axes (see diagram below).

Let the semi-major and minor axes be $a$ and $b$ respectively. Then we have:

$\begin{aligned} 5\left(\frac{a}{\cos{45^\circ}}\right)^2 + 5\left(-\frac{a}{\sin{45^\circ}}\right)^2 + 6 \left(\frac{a}{\cos{45^\circ}}\right)^2 \left(-\frac{a}{\sin{45^\circ}}\right)^2 & = 32 \\ 5\left(\frac{a}{\sqrt{2}}\right)^2 + 5\left(-\frac{a}{\sqrt{2}}\right)^2 + 6 \left(\frac{a}{\sqrt{2}} \right)^2 \left(-\frac{a}{\sqrt{2}}\right)^2 & = 32 \\ \frac{4a^2}{2} & = 32 \\ \Rightarrow a & = 4 \\ 5\left(\frac{b}{\cos{45^\circ}}\right)^2 + 5\left(\frac{b}{\sin{45^\circ}} \right)^2 + 6 \left(\frac{b}{\cos{45^\circ}}\right)^2 \left(\frac{b}{\sin{45^\circ}} \right)^2 & = 32 \\ 5\left(\frac{b}{\sqrt{2}}\right)^2 + 5\left(\frac{b}{\sqrt{2}} \right)^2 + 6 \left(\frac{b}{\sqrt{2}} \right)^2 \left(\frac{b}{\sqrt{2}}\right)^2 & = 32 \\ \frac{16b^2}{2} & = 32 \\ \Rightarrow b & = 2 \end{aligned}$

The area of the ellipse $A = \pi a b = 8 \pi \approx \boxed{25.133}$

link text
We see from above link that for an conic ellipse
$Ax^2+Bxy+Cy^2=1, \ \ area\ =\dfrac{2\pi}{\sqrt{4*AC-B^2}}\\$ $\therefore\ for\ 5x^2+6xy+5y^2=1,\ \ \ area\ =\dfrac{2\pi}{\sqrt{4*\frac{5*5}{32^2}-(\frac 6{32})^2}}=8\pi=25.1327$

Substitute $u=5x+3y$ , $v=4y$ , so that $u^2+v^2=5(5x^2+5y^2+6xy)$ . We get a circle $u^2+v^2=160$ with area $160\pi$ . The determinant of our transformation is $\begin{vmatrix} 5 & 3 \\ 0 & 4 \\ \end{vmatrix}=20$ so the area of the original curve is $8\pi$ .