Show that the area of the triangle does not depend on the location of the point. I'm having trouble with the following homework problem:

In a coordinate system $Oxy$ we have a variable rectangle $OABC$. Point $B$ is a point on the curve $K$ and can move along that curve. Independent from the position of $B$ we know that $OABC$ forms a rectangle with a constant area of $72$. $A$ is on the positive y-axis and $C$ on the positive x-axis.

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*Show that the function of curve $K$ is $y=\dfrac{72}{x}$.

The tangent to a point $B$ on the curve intersects the x-axis in $D$ and the y-axis in $E$.
2 . Show that the area of the triangle $ODE$ doesn't depend on the position of point $B$ on the curve $K$.


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*What I did: $72 = X_B \times Y_B$, so $y = \dfrac{72}{x}$.


*Here's where I don't exactly know what to do. We have a tangent to a certain point $ B$ with $(x, \dfrac{72}{x})$. But I don't know how to do this. I know that for 2 lines to be tangent in a certain point, their coordinates should be the same and the derivative in that point too, but I don't know how I can figure out the coordinates of $D$ and $E$ using this.
 A: For the equation $y=\dfrac{72}{x}$, we have $y'=-\dfrac{72}{x^2}$. Instead of using $\left(x,\dfrac{72}{x}\right)$ for $B$ (which is fine and works but sometimes causes confusion substituting the same variable on itself), I'm going to use $\left(b,\dfrac{72}{b}\right)$. Then the slope of the tangent line at $B$ is $-\dfrac{72}{b^2}$. Using point-slope form of the line $\left(y-y_0 = m(x-x_0)\right)$, we get the tangent line is
$y-\dfrac{72}{b} = -\dfrac{72}{b^2}(x-b)$
We only care about the $x$-intercept ($D$) and the $y$-intercept ($E$).
Since $D$ is the $x$-intercept, we must have $y=0$. Substitution into the tangent line equation yields
$0-\dfrac{72}{b} = -\dfrac{72}{b^2}(x-b) \\
\Longleftrightarrow -\dfrac{b^2}{72}\cdot-\dfrac{72}{b} = -\dfrac{b^2}{72}\cdot-\dfrac{72}{b^2}(x-b) \\
\Longleftrightarrow b = x-b \Longleftrightarrow x=2b$
So $D$ is $(2b,0)$.
Similarly for $E$, we must have $x=0$.
$y-\dfrac{72}{b} = -\dfrac{72}{b^2}(0-b)
\Longleftrightarrow y-\dfrac{72}{b} = \dfrac{72}{b}
\Longleftrightarrow y = \dfrac{144}{b}$
So $E$ is $\left(0,\dfrac{144}{b}\right)$.
The triangle $ODE$ is a right triangle with right angle at $O$. The lengths of the legs are just the non-zero coordinates of $D$ and $E$. So
$A_{\triangle} = \dfrac{1}{2}bh = \dfrac{1}{2}(2b)\left(\dfrac{144}{b}\right) = 144$. Since this answer is constant (does not contain $b$), the area of the triangle does not depend on where $B$ is located.
