How to find area of triangle in $1/x$ function? I'm trying to

Find a point $T[a, ?]$ on graph of function $f(x) = 1/x$, for which tangent line intersecting this point with line intersecting points $T;[0,0]$ form a triangle with area $S = 1$.

I'm aware that area has to be calculated using integrals, but I have no clue how to begin. I attempted to plot the problem and I got this plot. I understand that definite integrals are used to calculate content under the curve, but the triangle does not seem to follow the "under the curve" pattern and it confuses me a little. 
Can you get me started?
 A: So we look for a point $T$ of the form $(x,1/x)$ so we will study which value of $x$ gives the required triangle.
The height of the triangle will be $1/x$, so we need to find the size of its base which is a function of $x$.
From basic calculus we know that the equation of the tangent line at a point $(x_0,y_0)$ is $Y = y_0 + m(X -x_0)$ where $m$ is the slope.
Here our slope $m$ depends on $x$ and is $m = \left( 1/x \right)' = -1/x^2$, $\ x_0$ is our variable $x$ and $y_0$ is $1/x_0 = 1/x$.
Therefore the equation of the tangent (using capital $X,Y$) is
$$
Y = \frac{1}{x} - \frac{1}{x^2}(X - x)
$$
Which logically depends on $x$. We want to find where it intersect with the $X$-axis. Therefore we want $Y=0$ and solve for $X$ :
$$
\frac{1}{x} - \frac{1}{x^2}(X - x) = 0 \implies X = 2x
$$
Now we know that based on our choice of $x$, the tangent will cut the $x$-axis at $2x$ and therefore gives a base size of $2x$.
Applying now the $base \cdot height /2$ formula we get
$$
\frac{2x \cdot 1/x}{2} = 1
$$
Which is actually true for any $x \neq 0$ !
