Describe an equation of a parabola lying between (1,1) and the x-axis This question has me somewhat stumped. I need to do the following:
$$
\text{The parabola consisting of all points (x,y) that have the same }\\
\text{distance from the x-axis and the point (1,1) can be written as the graph of the equation}\\
y = \text{________. (Enter an expression in  x.)}
$$
I can only imagine this question means that I need to write the slope intercept form of a parabola that looks like this:

I have no idea how to do describe this, because (to my understanding) there isn't a mechanism to horizontally shift the slope intercept form.
The closest I can see actually getting to what is being asked for is 
$$
y = x^2 + \frac{1}{2}
$$
I could potentially see defining $m$ as something arbitrarily large to stretch the parabola, but even if that were the correct thing to do, how would one be able to pick a sufficiently large value that the line would be able to fullfill our requirements? (I don't imagine, we can). Any advice is greatly appreciated. Thank you. 
 A: The way the problem is described, the point $(1, 1)$ is the focus, and the $x$-axis is the directrix.
The easiest way to determine the equation for the resulting parabola, intuitively (in my opinion), is to identify three points: The apex of the parabola, which is exactly halfway in between the focus and the directrix, and the two points on either side of the focus but with the same $y$-coordinate.
Since the focus is at "height" $1$ above the directrix, the apex is at $(1, 1/2)$, and the two points on either side must be $1$ away in the $x$-coordinate: that is, at $(0, 1)$ and $(2, 1)$.  That way, they are equidistant from both the focus and the directrix.  We can therefore write
$$
y-\frac{1}{2} = k(x-1)^2
$$
This automatically places the apex at $(1, 1/2)$, and now we solve for the parabola that includes one of the other points, $(0, 1)$:
$$
\frac{1}{2} = k(1-0)^2 = k
$$
(The other point yields the identical result, as you can verify.)  Thus the equation for the parabola is
$$
y-\frac{1}{2} = \frac{(x-1)^2}{2}
$$
or, if you prefer,
$$
y = \frac{x^2}{2}-x+1
$$
A: Consider some point $(x,y)$ on the graph. Such a point must be the same distance from the $x$-axis as the point $(1,1)$. The distance from $(x,y)$ to $(1,1)$ is
$$\sqrt{(x-1)^2+(y-1)^2}$$
The closest point to the $x$-axis from $(x,y)$ is $(x,0)$, so the distance to the $x$-axis is simply the distance between these two points, which is simply $y$. Equating these distances, we get
$$\begin{align}
y&=\sqrt{(x-1)^2+(y-1)^2}\\
\implies y^2&=(x-1)^2+y^2-2y+1\\
\implies y&=\frac12(x-1)^2+\frac12=\frac12x^2-x+1
\end{align}$$
A: I believe you just want the equation of a parabola with focus $(1,1)$ and directrix $y=0$, which would be $y = \frac{x^2}{2}-x+1$.
We find this parabola by saying that since by definition a parabola is the locus of all points equidistant from a line and a fixed point, we know the distance from $(1,1)$ to some $(x_0,y_0)$ is $\sqrt{(x_0-1)^2+(y_0-1)^2}$. Furthermore the distance between $(x_0,y_0)$ and the directrix, $y=0$ is 
$|y_0 -0| = y_0$, by the aforementioned definition these expressions are equal: 
$$\sqrt{(x_0-1)^2+(y_0-1)^2} = y_0$$
$$\implies (x_0-1)^2+(y_0-1)^2 = y_0^2$$
$$y_0 = \frac{x_0^2}{2} -x_0 +1$$
Thus the desired parabola.
A: Notice, 
We have the coordinates of the vertex of parabola $\left(1, \frac{1}{2}\right)$ & focus at $(1, 1)$  
In general the equation of the parabola having vertex $(x_1, y_1)$ & having focus at $(x_1, a+y_1)$ is given as $$(x-x_1)^2=4a(y-y_1)$$
In this case, we have $$(x_1, y_1)\equiv \left(1, \frac{1}{2}\right)$$ & the focus at $(x_1, a+y_1)\equiv (1, 1)$ 
By comparing the corresponding coordinates, we get $x_1=1,\ y_1=\frac{1}{2}$ & $a+\frac{1}{2}=1\iff a=\frac{1}{2}$
Hence, the equation of the parabola is given as follows $$(x-1)^2=4\frac{1}{2}\left(y-\frac{1}{2}\right)$$ $$x^2-2x+1=2y-1$$ $$2y=x^2-2x+2$$
$$\implies y=\frac{1}{2}x^2-x+1$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=\frac{1}{2}x^2-x+1}}$$
