<s>[INST] The triangle formed by the $x$-axis, the tangent to $y=1/x$ at a 1st-quadrant point, and the line from the origin to that point is isosceles 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?
 [/INST] 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 $1$.
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