Here is Prob. 45, Sec. 3.2, in the book Calculus With Analytic Geometry by George F. Simmons, 2nd edition:
Let $P$ be a point on the first-quadrant part of the curve $y = 1/x$. Show that the triangle determined by the $x$-axis, the tangent at $P$, and the line from $P$ to the origin is isosceles, and find its area.
My Attempt:
Let the point $P$ be given by $P = \left( a, \frac{1}{a} \right)$, where $a > 0$.
Then the line from $P$ to the origin has equation $$ y = \frac{ \frac{1}{a} }{a} x = \frac{1}{a^2} x. \tag{1} $$
The slope of the tangent line to the curve $y = 1/x$ at the point $P$ is $$ \left( \frac{d y}{d x} \right)_{x = a} = -\frac{1}{a^2}, $$ and thus the equation of the tangent line is $$ y = -\frac{1}{a^2} (x-a) + \frac{1}{a} = -\frac{1}{a^2} x + \frac{2}{a}. \tag{2} $$ And, this line intersects the $x$-axis at the point $(x, 0)$, where $$ -\frac{1}{a^2} x + \frac{2}{a} = 0, $$ that is, $$ x = \frac{ \frac{2}{a} }{ \frac{1}{a^2} } = 2a. $$ Thus the tangent line to the curve at the point $(a, 1/a)$ intersects the $x$-axis at the point $A$ given by $$A = (2a, 0). $$
And, the point of intersection of the tangent line to the curve at point $P$ and the line from $P$ to the origin is of course the point $P = (a, 1/a)$ itself.
Thus the three vertices of our triangle are $O = (0, 0)$, $A = (2a, 0)$, and $P = (a, 1/a)$, where $a > 0$. So the sides of our triangle have lengths $$ \left| \overline{OA} \right| = 2a, $$ $$ \left| \overline{AP} \right| = \sqrt{ (2a-a)^2 + (0-1/a)^2 } = \sqrt{ a^2 + 1/a^2}, $$ and $$ \left| \overline{PO} \right| = \sqrt{ (a-0)^2 + (1/a-0)^2 } = \sqrt{a^2 + 1/a^2}. $$ Thus the sides $AP$ and $PO$ are congruent, showing that our triangle is indeed isosceles.
Finally, since the base of our triangle is the side $OA$ and the altitude is the vertical line segment from the $x$-axis to the point $P$, therefore the area of our triangle is $$ \frac{1}{2} \times (2a) \times \frac{1}{a} = 1. $$
Is what I have done correct and clear in each and every detail? Or, are there any issues of accuracy or clarity?
It looks good to me. A geometric perspective to consider:
When we find point $A =(2a, 0)$, notice that $P$ is directly above the midpoint $OA$. (This midpoint is $C = (a, 0)$.)
The triangles $OCP$ and $ACP$ are congruent, right-angled triangles that are mirror images of each other ($CP$ as a common edge).
Hence $OPA$ is an isosceles triangle.
Moreover if we rotate $ACP$ around $P$ until it forms a rectangle with $OCP$ (which occurs when $A$ moves to $O$), we get a rectangle with base $a$ and height $1/a$ hence it has area $1$.