I need to define a family (one parameter) of monotonic curves I want to define a function family $f_a(x)$ with a parameter $a$ in $(0,1)$, where:
For any $a$, $f_a(0) = Y_0$ and $f_a(X_0) = 0$ (see image)
For $a = 0.5$, this function is a straight line from $(0,Y_0)$ to $(X_0, 0)$.
For $a < 0.5$, up to zero (asymptotically perhaps), I want $f_a$ to be a curve below, and for $a > 0.5$, the curve should be to the other side. 
I didn't fill the diagram with many examples, but I hope you get the idea. Different values of $a$ always produce a distinct, monotonic curve, below all curves of larger values of $a$, and above all curves for smaller values of $a$. E.g.: when I decrease $a$, the distance of the $(0,0)$ point from the curve decreases, and if I increase $a$, it increases. 
Sorry for the clumsy description but I hope you got the intuition of what I'm trying to define! Any suggestion of how this function $f_a(x)$ could look like? 

 A: Here is a completely different approach that leads to much nicer behavior at the corners.
The idea is to take a line of negative unit slope, $y=\beta-x$, extending to infinity in both directions, and then squeeze it into your target rectangle using the logistic function and its inverse the logit.
The logistic function and logit contains exponentials and logarithm, but these mostly cancel out each other when we put the whole thing together, and we get
$$y = \frac{y_0}{1+e^{-\beta}\frac{x}{x_0-x}}$$
We then have to decide how $\beta$ must depend on $a$ -- your desired behavior will result if we set $e^{-\beta}=(1-a)/a$, to get the final definition
$$ y = \frac{y_0}{1+\frac{(1-a)x}{a(x_0-x)}}$$
This has a number of nice properties:


*

*It's a simple rational expression.

*For $a=1/2$ the curve is a straight line, as specified.

*The slope of the curve varies smoothly with $a$ everywhere, even at the endpoints.

*When $x_0=y_0$, all curves are symmetric about the line $x=y$.

*When $x_0=y_0$, replacing $a$ with $1-a$ will simply reflect the curve about the diagonal.

A: Perhaps $x^n + y^n = 1$, or $y = (1-x^n)^{1/n}$, will serve?
           


Above I set $n=\frac{1}{4},\frac{1}{3},\frac{1}{2} ,1,2,3,4$.
Then adjust for your $x_0$ and $y_0$.
A: I'd recommend putting everything the others said together:
$$f_a(x)=Y_0\left(1-(x/X_0)^{a/(1-a)}\right)^{(1-a)/a} $$
This way you 


*

*Get the desired curve at $a=1/2$ (as in Makholm and Israel's answers) and

*Get the desired curves as $a\to0$ or $1$ and 

*Get the curves stretched to the desired rectangle framed by $X_0$ and $Y_0$ and

*Get the curves to look like as they do in Joseph's answer (or a previous answer of mine), i.e. they are symmetric about the line $y=x$ before being stretched to the rectangle.


Low-res colored examples (taken from stretching a graphic from Mathworld, see link):
$\hskip 1.5in $ 
A: You might try $f_a(x) = y_0 (1 - x/x_0)^{(1-a)/a}$.
A: How about
$$f_a(x) = y_0\left(1-(x/x_0)^{\frac{a}{1-a}}\right)$$
?
