Directrix and focus of $ax^2 + bx + c$

Question

How can you find the directrix and focus of a parabola (quadratic function) $$ax^2 + bx + c,$$ where $a \neq 0?$ I mean, given the focus $x, y$ and directrix (I'll use a horizontal line for simplicity) $y = k$ you can find the equation of the quadratic; how do you do this backwards?

Answer 1

The goal is essentially to get $y = ax^2 + bx + c$ into the form $4p(y - k) = (x - h)^2$. This can be done by completing the square:

\begin{align} y &= ax^2 + bx + c \\ \frac{y - c}{a} &= x^2 + \frac{b}{a} x \\ \frac{y - c}{a} + \left(\frac{b}{2a} \right)^2 &= x^2 + \frac{b}{a}x + \left(\frac{b}{2a} \right)^2 \\ \frac{1}{a} \left(y - c + \frac{b^2}{4a} \right) &= \left(x + \frac{b}{2a} \right)^2. \end{align}

Thus the parabola can be written as

$$ 4 \cdot \frac{1}{4a} \left(y - \left(c - \frac{b^2}{4a} \right) \right) = \left(x - \left(-\frac{b}{2a} \right) \right)^2. $$

The focus is

$$ \left(-\frac{b}{2a}, c - \frac{b^2}{4a} + \frac{1}{4a} \right) = \left(-\frac{b}{2a}, \frac{1 - b^2}{4a} + c \right) $$

and the directrix is

$$ y = c - \frac{b^2}{4a} - \frac{1}{4a} = c - \frac{1 + b^2}{4a}. $$

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Update: Why $4p(y - k) = (x - h)^2$ is a useful form.

Starting from scratch, suppose we want to construct an upward or downward opening parabola (i.e., one of the form $y = ax^2 + bx + c$). Let its vertex be $(h, k)$. The definition of a parabola is that it is the set of all points equidistant to a point (the focus) and a line (the directrix). So let $(h, k + p)$ be the focus and $y = k - p$ be the directrix for some $p$. (Note that the vertex $(h, k)$ is midway between the focus and directrix, as required.)

Let $(x, y)$ be a point on the parabola. This means that its distance to the point $(h, k + p)$ is equal to its (shortest) distance to the line $y = k - p$. The point $(x, k - p)$ on the directrix is the closest point to $(x, y)$ (draw a picture to see why). So we equate distances:

\begin{align} \sqrt{(x - h)^2 + (y - (k + p))^2} &= \sqrt{(x - x)^2 + (y - (k - p))^2} \\ (x - h)^2 + (y^2 - 2(k + p)y + (k + p)^2) &= y^2 - 2(k - p)y + (k - p)^2 \\ (x - h)^2 + (y^2 - 2ky - 2py + k^2 + 2kp + p^2) &= y^2 - 2ky + 2py + k^2 - 2kp + p^2 \\ (x - h)^2 &= 4py - 4kp \\ &= 4p(y - k). \end{align}

Therefore, if we have a parabola with equation $4p(y - k) = (x - h)^2$, its focus will be $(h, k + p)$ and its directrix will be $y = k - p$.

Answer 2

Okay, so for to answer your question I started off with making a standard parabola from the given equation.

$$y= ax^2+bx+c$$ $$y-c+\frac{b^2}{4a^2}= ax^2+2(\frac{ba}{2a})x+\frac{b^2}{4a^2}$$ $$(ax +\frac{b}{2a})^2=y-c+\frac{b^2}{4a^2}$$ $$(x +\frac{b}{2a^2})^2=\frac{1}{a^2}(y-c+\frac{b^2}{4a^2})$$

This gives you a standard equation in form of
$$(x-H)^2=4A(y-B)$$

Thus it's clear that your Vertex of this parabola are : $$(\frac{-b}{2a^2},c+\frac{-b^2}{4a^2})$$

Now, Let an equation be $$X^2=4Y $$ where $$X=x-H$$ and $$Y=y-B$$ for this parabola, A=1 Hence it's focus is $$(0,-1)$$ implying $$X=0$$ and $$Y=-1$$ to get the actual focus, simply substitute X and Y respectivel according to values given to get actual focus on $$(H,B-1)$$ where H and B are already specified. Now to get your directrix,put $$Y-1=0$$ to get actual directrix as$$y=B+1$$ I hope it solves your query. Cheers!