Area enclosed by the graph of $13x^2-20xy+52y^2+52y-10x=563$. Find the area enclosed by the graph of $13x^2-20xy+52y^2+52y-10x=563$.
First I saw that this cannot be a circle ($xy$ term), and it cannot be an ellipse with axes parallel to the coordinate axes. But its obviously a closed figure. I tried to factor it as a sum of squares equal to zero, but this did not work. By plugging into W|A I find that it is an ellipse with sides not parallel to the coordinate axes, but centered at the origin. First, I have no idea to deduce that simply by manipulating the equation algebraically (unless you know the general form of ellipses not necessarily parallel to the coordinate axes). I hope someone can tell without giving an ad hoc explanation. Second, even knowing that the figure is an ellipse, I still don't know how to determine the minor and major axes (but from there its easy by using area equals $\pi a b$).
 A: We can write the quadratic part as follows:
$$13x^2 - 20xy + 52y^2 = \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}13 & -10 \\ -10 & 52\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$$
By choosing a change of coordinates that diagonalizes this matrix, we can transform any equation like this one into an ellipse with axes parallel to the coordinate axes.

We can find a solution as follows: set $x' = x-a$, $y'=y-b$, plug in and choose $a,b$ so that the linear terms vanish.  As Jonas mentions in his comment, this gives us $(a,b) = (0,1/2)$, and the equation:
$$13x'^2 -20x'y' + 52y^2 = 576$$
The equation $x''^2 + y''^2 = 576$ is a circle of area $576\pi$.  So the above equation bounds a region of area $576\pi / \sqrt{\operatorname{det} A} = 576\pi / 24 = 24\pi$.
More concretely (and with less linear algebra required), we can write $ux^2 + vxy + wy^2 = u(x+\frac{v}{2u}y)^2 + (w - \frac{v^2}{4u})y^2$, so we can compute the scaling factor directly as $\sqrt{u(w-\frac{v^2}{4u})} = \sqrt{uw-\frac{v^2}{4}}$ (in this case, $24$).
A: Note that you can rewrite this, with a little work, as $$4(x+2y+1)^2+9(-x+2y+1)^2=576=24^2$$
If we put $X=x+2y$ and $Y=-x+2y$ the determinant of the transformation is $4$ and we get $$\frac {(X+1)^2}{12^2}+\frac {(Y+1)^2}{8^2}=1$$ as a transformed ellipse with its axes parallel to the new co-ordinate axes. The new ellipse has major/minor axes of lengths $12$ and $8$ and area $\pi\times 12\times 8=96\pi$. The original ellipse has area one quarter of this i.e. $24\pi$.

How to rewrite - I just hacked it out by equating coefficients with the form $(ax+by+c)^2+(dx+ey+f)^2$. This took me less than ten minutes on paper, and is definitely elementary

It would also have been possible to shift the origin first to get the symmetric form $$13x^2-20xy+56y^2=576$$Then you want $$(ax+by)^2+(cx+dy)^2=13x^2-20xy+56y^2$$
So $a^2+c^2=13, b^2+d^2=56, ab+cd=-10$
So go for $a=\pm 2$ and $c=\pm 3$ and spot $2\cdot 4-3\cdot 6=-10$ so that we can rewrite as $$(2x+4y)^2+(-3x+6y)^2$$ which is essentially what I did, and the arithmetic from there is essentially the same.

On the determinant of the transformation, that's simply a matrix determinant. We write $$\begin{bmatrix}X \\ Y\end{bmatrix}=\begin{bmatrix}1 & 2 \\ -1 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$$
The determinant of the matrix is $1\times 2-2\times -1=4$. The determinant in two dimensions gives you the scale factor for area (in three dimensions it would be volume, etc).
A: If you look closely at W|A's graph you will see that it is not actually centered at the origin, just close to it. If it were centered at the origin it would be symmetric with respect to the origin, and replacing $x$ with $-x$ and $y$ with $-y$ would give the same equation, but it does not.
I can see two ways to solve this problem, non-ad-hoc. First, do the rotation that @ADG gave in his first answer. (That technique uses a change of variables to rotate the axes by the angle $\theta$ that satisifies $\tan 2\theta=\frac{B}{A-C}=\frac{-20}{13-52}=\frac{20}{39}$). As @ADG said, that will be messy, but it will definitely work (if you don't make a mistake).
Another way is to use the quadratic formula to solve for $y$ in terms of $x$. This will give you a range of possible values of $x$ and two values of $y$ for such $x$'s, a lower and an upper one. You could do an integral of the difference of $y$ values between the limiting $x$ values. There would be some cancellation in the difference of $y$ values, so you end up doing the integral of the square root of a quadratic polynomial. In advance this seems to be difficult, but there seems to be enough cancellation that this may not be a bad method.
If it helps, you could use the discriminant of the equation to see if it is an ellipse. In your case the discriminant is $(-20)^2-4*13*52=-2304$. This is negative, so the graph is indeed an ellipse. This is standard and not ad hoc.
By the way, W|A says that the area is $24\pi$ and appears to use my second suggestion, using the quadratic formula and integration.

Here are some details for the quadratic formula and integration approach:
Rearrange your equation to get
$$52y^2+(-20x+52)y+(13x^2-10x-563)=0$$
Solving this for $y$ using the quadratic formula and simplifying,
$$y=\frac{5x-13\pm12\sqrt{52-x^2}}{26}$$
Clearly $x$ runs from $-\sqrt{52}$ to $\sqrt{52}$. The area we want is
$$\int_{-\sqrt{52}}^{\sqrt{52}}(y_2-y_1)\,dx$$
$$=\int_{-\sqrt{52}}^{\sqrt{52}}2\frac{12\sqrt{52-x^2}}{26}\,dx$$
$$=\frac{6}{13}\int_{-\sqrt{52}}^{\sqrt{52}}2\sqrt{52-x^2}\,dx$$
The integral is the area of the circle with radius $\sqrt{52}$, so we get
$$\frac{6}{13}\cdot 52\pi$$
$$=24\pi$$
There, that wasn't so bad! (Though I made a mistake when I first did the calculation and typed it in.)
A: First of all, the curve is clearly a non-degenerate conic or else we won't be finding its area. In the equation of a conic if the coefficient of $xy$ is $2h$, of $x^2$ is $p$ and of $y^2 $ is $q$ then the rule is that if $h^2<pq$, then it is an ellipse, which it is.
The centre of a conic represented by the equation $ax^2+by^2+2hxy+2gx+2fy+c=0$ is given by $C\equiv \left(\dfrac{hf-bg}{ab-h^2},\dfrac{hg-af}{ab-h^2}\right)$ which in this case is $C\equiv\left(0,\dfrac{-1}{2}\right)$

Proof:Let the cenre be at $(x_0,y_0)$. Then the equation of diameter parallel to the X-axis is $y=y_0$, and substituting that in the equation of conic we get $ax^2+(2hy_0+2g)x+by_0^2+2fy_0+c=0$. If $x_1$ and $x_2$ are the roots of this equation then the centre's X-coordinate is $$x_0={x_1+x_2 \over 2}={-g-hy_0\over a}\implies ax_0+hy_0+g=0\tag{i}$$
  A similar procedure with the diameter $x=x_0$ gives $$by_0+hx_0+f=0\tag{ii}$$
  Solving (i) and (ii) gives the result.

Now take any rational point on the ellipse, say, $P\equiv\left(4,\dfrac{-5}{2}\right)$.
The equation of the diameter $CP$ is $x+2y+1=0$.

A useful property of ellipses:The tangent at the endpoint of a diameter is parallel to its conjugate diameter.

The equation of the tangent at point $P$ is found by the standard method, and the slope of the tangent works out as $\dfrac{1}{2}$, which should also be the slope of the diameter conjugate to $CP\implies$ the equation of conjugate diameter $CD$ is $x-2y-1=0$.
Taking intersection of the line $CD$ and the ellipse you get point $D\equiv\left(6,\dfrac{5}{2}\right)$.
The equation of normal at $P$ is found easily and is $4x+2y-11=0$. The foot of perpendicular from the centre $C$ to the normal is simply the intersection of the normal and line $CD\implies$ $F\equiv\left(\dfrac{12}{5},\dfrac{7}{10}\right)$


Another useful property of ellipses: Let $F$ be the foot of the perpendicular drawn from the centre $C$ to the normal drawn at point $P$ on the ellipse. Let $D$ be the endpoint of the diameter conjugate to the diameter $CP$. Then $PF.CD$ is equal to the product of the lengths of the semi axes.

Above stated property of ellipse says that $PF.CD=ab$. Now just use the distance formula . $PF$ works out to be $\dfrac{8}{\sqrt5}$ and $CD$ works out as $3\sqrt5$. This gives the area=$\color{red}{24\pi}$ 

Proof of the property: Consider the the standard ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and the point $P$ on it with eccentric angle $\theta$ and draw whatever is required. The equation of the tangent at $P$ is $bx\cos\theta+ay\sin\theta=ab$. Since the distance $PF$ is simply the distance of the tangent from origin it is easily computed to be $\dfrac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}$. Since the point $D$ is at the endpoint of a semi-conjugate diameter, its eccentric angle is $\dfrac{\pi}{2}+\theta$, so the distance $CD$ is simply $\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$. Now just multiply the two. The square root term cancels and this completes the proof.

A: 
$$13x^2-20xy+52y^2+52y-10x=563$$

As $(-20/2)^2<13.52$ and $$\left|\begin{matrix}13&-20/2&-10/2\\-20/2&52&52/2\\-10/2&52/2&-563\end{matrix}\right|<0$$
Both of which iff "the equation is of an ellipse"
By partial differentiation center is $(0,-1/2)$, i.e. solving $26x-20y-10=0,-20x+104y+52=0$. After shifting our origin to that point, we get using polar coordinates (all terms of $x,y$ will be eliminated.) and using double angle formulae:
$$r^2=\frac{576}{13\cos^2\theta+52\sin^2\theta-20\sin\theta\cos\theta}=\frac{2.24^2}{65-(39\sin2\theta-20\sin2\theta)}$$
Using the fact that $|a\cos\theta+b\sin\theta|\le\sqrt{a^2+b^2}$ we can get:
$$r_{\rm max}=\frac{24\sqrt2}{\sqrt{65-\sqrt{1921}}},r_{\rm min}=\frac{24\sqrt2}{\sqrt{65+\sqrt{1921}}}\implies A=\pi r_{\rm max}r_{\rm min}=\pi\frac{24^2.2}{\sqrt{65^2-1921}}=24\pi$$
A: Area of ellipse given by the equation$$ax^2+2hxy+by^2+2gx+2fy+c=0$$ where $$\triangle=abc+2fgh-af^2-bg^2-ch^2\neq 0$$ and $$h^2<ab$$
is given by
$$\frac{\pi (\triangle)}{(ab-h^2)^{3/2}}$$
A: I really like the variety of methods included here. I will summarize them and also add a new method.
It is a neat fact (that I was previously unaware of) that conic sections are quadratic polynomials in $x$ and $y$. Once you realize that the equation describes a conics, the problem implies the conic is closed, thus its an ellipse. Furthermore due to the $xy$ term the ellipse is oblique. Also note that the ellipse is not centered at the origin. There are a few ways to proceed.
Analytic:


*

*Center the ellipse at the origin by making the appropriate substitutions so that the $x$ and $y$ terms disappear. ADG proposes that we find convert to polar coordinates and find the maximum and minimum values of $r$ (with calculus or other methods) which give the semimajor and semiminor axes respectively, and hence the area.

*Solve the equation for $y$ in terms of $x$ as Daulton suggests, using the quadratic formula. Find the horizontal boundaries of the parabola by using the discriminant.Then find the area by integrating between these two boundaries. The integral is equivalent to finding the area of a semicircle so this is totally doable. We don't have to bother really with centering the ellipse at the origin as the calculations are just as annoying computationally.
Matrices:
3.Find the rotation matrix and relate the area to that of a circle via shear mapping.
Other:


*Center at origin. Then rewrite the equation as a sum of squares and make a substitution to transform to the standard form of an ellipse centered at the origin with axes parallel to the coordinate axes. Then use the determinant of the transformation.

*Do what G-man did with geometric properties of ellipses.

Another method:
First center the ellipse so as to make the $x$ and $y$ terms disappear. We get something of the form 
$Ax^2+Bxy+Cy^2=D$
We will rotate the ellipse so its parallel to the coordinate axes, and we do this by removing the $xy$ term. Note that this does not change the discriminant or the constant term (why?).  First divide by $D$:
$(A/D)x^2+(B/D)xy+(C/D)y^2=1$
Suppose after the rotation the equation is 
$(1/P^2)x^2+(1/Q^2)y^2=1$
This is the standard form of an ellipse with semimajor axis $P$ and semiminor axes $Q$, so the area is $\pi PQ$. Thus it suffices to find $PQ$.
We can calculate the discriminant $\lambda=(B/D)^2-4(A/D)(C/D)$ and set this equal to $-4(1/P^2)(1/Q^2)$ to solve for $PQ$. 
