Rectangle area and a curve The diagonals of a rectangle are both 10 and intersect at (0,0). Calculate the area of this rectangle, knowing that all of its vertices belong to the curve $y=\frac{12}{x}$.
At first I thought it would be easy - a rectanlge with vertices of (-a, b), (a, b), (-a, -b) and (a, -b). However, as I spotted no mention about the rectangle sides being perpendicular to the axises, it's obviously wrong which caused me to get stuck. I thought that maybe we could move in a similar way - we know that if a rectangle is somehow rotated (and we need to take that into account), the distances from the Y axis from the points being symmetric to (0,0) are still just two variables. So we would have: (-b, -12/b), (a, 12/a), (-a, -12/a), (b, 12/b). I then tried to calculate the distance between the first two and the second and the third which I could then use along with the Pythagorean theorem and a diagonal. However, the distance between the first two is $\sqrt{(a+b)^2+(\frac{12}{a}+\frac{12}{b})^2}$ which is unfriendly enough to make me thing it's a wrong way. Could you please help me?
 A: To follow J.M.'s hint, you need to solve the two equations
$$
\eqalign{
x^2+y^2&=25\\
xy&=12
}
$$
and a nice way to do this would be to notice that then
$$
\left(x+y\right)^2=x^2+2xy+y^2=25+2\cdot12=49
$$
so that $x+y=\pm7$. Next, if you note that
$$
\left(x-y\right)^2=x^2-2xy+y^2=25-2\cdot12=1
$$
you will see that $x-y=\pm1$. This gives us the system
$$
\eqalign{
x+y&=\pm7\\
x-y&=\pm1
}
$$
From which we can add or subtract, then divide by two, to get
$$
x,y=\pm\frac{7\pm1}{2}=\pm3,\pm4
$$
from which we can work out the combination of signs
(in the first and third quadrants) that satisfy the
equations above:
$$
\{(3,4),(4,3),(-3,-4),(-4,-3)\}
$$
Since the (within- and between-quadrant) sides of
the rectangle are $\sqrt{1^2+1^2}=\sqrt2$ and
$\sqrt{7^2+7^2}=7\sqrt2$, the area is their product,
$$A = \sqrt2\cdot7\sqrt2 = 14\,.$$
A: As in J.M.'s comment, the diagonals of a rectangle (any parallelogram, in fact) bisect each other, so we're looking for points on $y=\frac{12}{x}$ that are a distance of $5$ from the origin.  That is, we want solutions to the system $$\left\{\begin{matrix}
y=\frac{12}{x}\\ 
x^2+y^2=5^2
\end{matrix}\right..$$  By substituting for $y$ in the second equation, $$\begin{align}
&&x^2+\left(\frac{12}{x}\right)^2&=25 \\
&\implies&x^2+\frac{144}{x^2}&=25 \\
&\implies&x^4+144&=25x^2 \\
&\implies&x^4-25x^2+144&=0 \\
&\implies&(x^2-9)(x^2-16)&=0 \\
&\implies&(x-3)(x+3)(x-4)(x+4)&=0 \\
&\implies&x=\pm3\text{ or }\pm4
\end{align}$$
Using $y=\frac{12}{x}$ gives the corresponding $y$-coordinates for each of the 4 points, from which the side lengths and area can be computed.
