The function $f(x),$ defined for $x > 0,$ is positive, differentiable, and decreasing. The function $f(x)$ has the following property: Let $P$ be a point on the graph, and draw the tangent to the graph at $P.$ Let the tangent meet the $x$-axis and $y$-axis at $A$ and $B,$ respectively. Then the length of chord $AB$ is constant.
Find all functions $f(x)$ that have this property.
Setting $x=0,$ we obtain
\[0 = f(0) = (1 - 0) f'(0),\]which means $f(0) = 0.$
[asy]
unitsize(1 cm);
real func (real x) {
return(exp(-x^2));
}
pair T(real x) {
real dy = 2*x*exp(-x^2);
real dx = 1;
real m = dy/dx;
return((x - m,x + m));
}
draw(graph(func,0.03,1.3));
draw((0,0)--(0,1),dashed);
draw(T(0));
draw(T(0.1));
draw(T(0.2));
draw(T(0.3));
draw(T(0.4));
draw(T(0.5));
draw(T(0.6));
draw(T(0.7));
draw(T(0.8));
draw(T(0.9));
draw(T(1));
draw(T(1.1));
draw(T(1.2));
draw((0,0)--(1.7,0),dashed);
label("$A$", (0,0), W);
label("$B$", (0,func(0)), NW);
label("$T$", (0.8,0), S, UnFill);
[/asy]
Let $T$ be the point of tangency. Then by the Two Chord Theorem, $AB$ pass through a fixed point $T.$ The slope of $AB$ is equal to the slope of the tangent line at $T.$
Hence, all the tangent lines are concurrent, so the graph of $f(x)$ must approach a horizontal line $L$ as $x \to \infty.$
Let $k = f'(x).$ Let $T_x$ be the point on the graph where the tangent line meets the $x$-axis, and let $B_x = (0,f(x))$ be the point where the tangent line meets the $y$-axis. Then by the Two Chord Theorem, all the lines $T_x B_x$ pass through a point $T$ on the graph.
Let $y = f(x)$ and $x = g(y).$ Then the slope of $\overline{T_x B_x}$ is
\[\frac{f(x) - 0}{0 - g(f(x))} = -\frac{g(y)}{g'(y)} = -\frac{x}{f'(x)},\]so
\[\frac{dy}{dx} = -\frac{x}{f'(x)}.\]By the Product Rule,
\begin{align*}
-f'(x) + xf''(x) &= 0, \\
f''(x) &= \frac{f'(x)}{x}.
\end{align*}Let $u = f'(x).$ Then $u = \frac{f(x)}{x},$ so
\[u' = \frac{xf'(x) - f(x)}{x^2}.\]Then using the fact that $f'(x) < 0,$ $f(x) > 0,$ and $f(0) = f'(0) = 0,$ we find
\begin{align*}
u' &= \frac{xf'(x) - f(x)}{x^2} \\
&= \frac{xf'(x) - x f(x)/x}{x^2} \\
&= \frac{x f'(x) - x f'(x)/x}{x^2} \\
&= 0.
\end{align*}Therefore, $u = C$ for some constant $C.$ Hence,
\[f'(x) = Cx.\]If $C = 0,$ then $f'(x) = 0$ for all $x,$ which means $f(x)$ is constant. But then $f(x)$ is not positive.
Otherwise, let $C \neq 0.$ Then we can write
\[f'(x) = Cx\]as $\frac{f'(x)}{f(x)} = Cx.$ Integrating both sides, we get
\[\ln |f(x)| = \frac{1}{2} Cx^2 + D\]for some constant $D.$ Then
\[|f(x)| = e^D e^{\frac{1}{2} Cx^2}.\]Since $f(x) > 0,$ $e^D$ is positive. We can absorb $e^D$ into the constant $E,$ so
\[f(x) = E e^{\frac{1}{2} Cx^2}.\]Since $f(0) = 0,$ $E = 0,$ so $\boxed{f(x) = 0}$ is the only solution.
Let's analyze the problem step by step.
Step 1: Finding the equation of the tangent line
Given a point $P = (x,y)$ on the graph of $f(x)$, the slope of the tangent line at $P$ is given by $f'(x)$. Using the point-slope form of a line, the equation of the tangent line passing through $P$ is:
$y - y_1 = m(x - x_1)$
where $m = f'(x)$ and $(x_1, y_1)$ are the coordinates of $P$. Rearranging the equation, we have:
$y = f'(x)(x - x_1) + y_1$
Step 2: Finding the coordinates of $A$ and $B$
For a given point $P = (x,y)$ on the graph, the x-coordinate of point $A$ is the root of the tangent line equation, i.e., $x$ such that $y = 0$. Substituting $y = 0$ into the equation of the tangent line, we have:
$0 = f'(x)(x - x_1) + y_1$
Solving for $x$, we get:
$x = x_1 - \frac{y_1}{f'(x)}$
Point $A$ has coordinates $(x, 0)$.
Similarly, for point $B$, the y-coordinate is the root of the tangent line equation, i.e., $x = 0$. Substituting $x = 0$ into the equation of the tangent line, we have:
$y = f'(x)(- x_1) + y_1 = -f'(x)x_1 + y_1$
Point $B$ has coordinates $(0, -f'(x)x_1 + y_1)$.
Step 3: Finding the length of chord AB
The length of chord $AB$ can be found using the distance formula. The length between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Substituting the coordinates of points $A$ and $B$ into the distance formula, we have:
$AB = \sqrt{(x - 0)^2 + (0 - (-f'(x)x_1 + y_1))^2} = \sqrt{x^2 + (-f'(x)x_1 + y_1)^2}$
Step 4: Analyzing the length of chord AB
According to the problem, the length of chord AB is constant for any point $P$ on the graph. This means that the expression inside the square root in the distance formula, $\sqrt{x^2 + (-f'(x)x_1 + y_1)^2}$, is constant. Let's denote this constant value as $C$:
$\sqrt{x^2 + (-f'(x)x_1 + y_1)^2} = C$
Squaring both sides of this equation, we get:
$x^2 + (-f'(x)x_1 + y_1)^2 = C^2$
Expanding and rearranging the equation, we have:
$x^2 + f'(x)^2x_1^2 - 2f'(x)x_1y_1 + y_1^2 = C^2$
Since the above equation should hold for any point $P$, the coefficients of $x^2$ and $f'(x)^2x_1^2$ must be equal to zero:
$1 = 0$ and $1 = 0$
These equations are contradictory, which means that no function $f(x)$ exists that satisfies the given property.
Therefore, there are no functions $f(x)$ that have the specified property.