1
$\begingroup$

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:

hand drawn image of a parabola

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.

$\endgroup$
1
  • $\begingroup$ What do you consider to be the "slope intercept form" of a parabola? There is an "intercept" form (really an x-intercept form) that is very easily shifted horizontally--just change the intercepts. But this parabola has no x-intercepts so it is not that simple. $\endgroup$
    – David K
    Aug 8, 2015 at 4:30

4 Answers 4

1
$\begingroup$

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,

$$ y = \frac{x^2}{2}-x+1 $$

$\endgroup$
1
$\begingroup$

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 $$\begin{align} y&=\sqrt{(x-1)^2+(y-1)^2}\\ \implies y^2&=(x-1)^2+y^2-2y+1\\ \implies y&=\frac12(x-1)^2+\frac12=\frac12x^2-x+1 \end{align}$$

$\endgroup$
1
  • $\begingroup$ check this calculation at the end $\endgroup$
    – Mr.Fry
    Aug 8, 2015 at 2:59
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $$(x-1)^2=4\frac{1}{2}\left(y-\frac{1}{2}\right)$$ $$x^2-2x+1=2y-1$$ $$2y=x^2-2x+2$$ $$\implies y=\frac{1}{2}x^2-x+1$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=\frac{1}{2}x^2-x+1}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.