How to set up a formula to get equations of different parabolas but sharing the same focus?

Question

this is my first question so I hope I got it right. Well I am trying to change the design of a parabolic solar concentrator and I was wondering if it was possible to have different equations of parabolas for the same focus (of the type $y=ax^2+bx+c$), since in my design I would like to keep the focus fixed. With the attached image, I think my point is better understood, you can see 3 parabolas, one more open than others, however all 3 have the same focus (I drew the red ones, so they have no equation). What I am looking for is that, to draw different parabolas modifying some parameter, but maintaining a focus that I initially set. That's all, I hope I have expressed my point well and if someone knows how I could get it, I would appreciate it, since I still have not achieved it.

Answer 1

This will be a quite straightforward solution, and condsider it is not unique (I think).

The focus of a parabola of the type $y = ax^2 + bx + c$ has coordinates

$$F = \left(-\frac{b}{2a};\ \frac{1-\Delta}{4a}\right)$$

Where $\Delta = b^2-4ac$.

Now let's say you want a focus of coordinates (chosen by you) $F = (m, n)$ that is you need to have

$$ \begin{cases} -\dfrac{b}{2a} = m \\\\ \dfrac{1 - \Delta}{4a} = n \end{cases} $$

With a bit of mathematics, they become

$$ \begin{cases} b = -2am \\\\ 4a^2m^2 - 4an - 4ac - 1 = 0 \end{cases} $$

Now remember that you CHOSE the values for $m$ and $n$ so they are fixed numbers. The inverse problem is to find all the possible values for $a, b, c$ that satisfy the equations above to give you those values of $m$ and $n$, that is: the same focus for every parabola.

This can be done now by solving the second equations for $a$, getting

$$a = \frac{(n+c)\pm\sqrt{(n+c)^2 + m^2}}{2m^2}$$

and getting back to $b$ you obtain

$$b = -\frac{(n+c)\pm\sqrt{(n+c)^2 + m^2}}{m}$$

Now as I said, $m, n$ are numbers you chose from the beginning. You just need to pick some value for $c$ that satisfies you, and you're done.

So eventually, choosing two fixed numbers $m, n$ and varying the value of $c$, the focus of a parabola is $F = (m, n)$ and all the parabolas of the form

$$y = \frac{(n+c)\pm\sqrt{(n+c)^2 + m^2}}{2m^2} x^2 -\frac{(n+c)\pm\sqrt{(n+c)^2 + m^2}}{m} x + c$$

will have the same foci of coordinates $F = (m, n)$.

Answer 2

The focus-directrix property of parabolas tells us that, for any point $P$ on the curve, $$\text{distance from $P$ to the focus} \;=\; \text{distance from $P$ to the directrix} \tag1$$

Let the parabola have a fixed focus $F=(m,n)$. If the parabola's axis of symmetry is presumed vertical, then its directrix is a horizontal line; say $y=d$. Writing $P=(x,y)$, the distances to $F$ and the directrix are easy to compute, so that $(1)$ becomes $$\sqrt{(x-m)^2+(y-n)^2} \;=\; |y-d| \tag2$$

Squaring, gathering terms, and isolating $y$ gives

$$y = \frac{1}{2(n-d)}\left(\,(x-m)^2+n^2-d^2\,\right) \tag{$\star$}$$

(This becomes @Lapacian's answer (replacing "$\pm$" with "$+$") by taking $d=-c+\sqrt{(n+c)^2+m^2}$.)