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### Course: Precalculus (2018 edition)>Unit 2

Lesson 7: Focus and directrix of a parabola

# Parabola focus & directrix review

Review your knowledge of the focus and directrix of parabolas.

## What are the focus and directrix of a parabola?

Parabolas are commonly known as the graphs of quadratic functions. They can also be viewed as the set of all points whose distance from a certain point (the focus) is equal to their distance from a certain line (the directrix).

## Parabola equation from focus and directrix

Given the focus and the directrix of a parabola, we can find the parabola's equation. Consider, for example, the parabola whose focus is at $\left(-2,5\right)$ and directrix is $y=3$. We start by assuming a general point on the parabola $\left(x,y\right)$.
Using the distance formula, we find that the distance between $\left(x,y\right)$ and the focus $\left(-2,5\right)$ is $\sqrt{\left(x+2{\right)}^{2}+\left(y-5{\right)}^{2}}$, and the distance between $\left(x,y\right)$ and the directrix $y=3$ is $\sqrt{\left(y-3{\right)}^{2}}$. On the parabola, these distances are equal:
$\begin{array}{rl}\sqrt{\left(y-3{\right)}^{2}}& =\sqrt{\left(x+2{\right)}^{2}+\left(y-5{\right)}^{2}}\\ \\ \left(y-3{\right)}^{2}& =\left(x+2{\right)}^{2}+\left(y-5{\right)}^{2}\\ \\ {y}^{2}-6y+9& =\left(x+2{\right)}^{2}+{y}^{2}-10y+25\\ \\ -6y+10y& =\left(x+2{\right)}^{2}+25-9\\ \\ 4y& =\left(x+2{\right)}^{2}+16\\ \\ y& =\frac{\left(x+2{\right)}^{2}}{4}+4\end{array}$

Problem 1
Write the equation for a parabola with a focus at $\left(6,-4\right)$ and a directrix at $y=-7$.
$y=$

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• The situation where you are given, for example x=4 instead of y=4, was never covered in the videos.
• Another consideration would be that when the directrix is a vertical line (x=k), we are representing a parable that is faced to the right or to the left which is no longer a function unless its domain is limited to represent only one "arm" of the parable.
• couldn't you use the equation y= a(x-h)^2 +k and x=a(y-k)^2 +h, where a=1/4p?
• In the equations y=a(x-h)^2+k and x=a(y-k)^2+h where a=1/4p , (h,k ) represents vertex and p is the distance between focus and vertex .
But in the equations y=1/2(b-k) (x-a)^2+ (b+k)/2 and x=1/2(a-k ) (y-b)^2 +(a+k)/2
In the above equations (a,b) represents FOCUS not VERTEX . In the first one directrix is y=k and in second one directrix is x=k . In the equations given by you in the question directrix in the first one is y=k-p and in second one it is x=h-p.
• Why did you factor (y-5)^2 but not (x+2)^2 ?
in a problem in Khan Academy, I factored X but I got the wrong answer! the right answer was to factor the Y !!
I don't understand!
here's the problem:
focus at (2,2), directrix x=8
• If you need to find the X you multiply out the (x+2)^2.If you need to find the Y you factor out (y-5)^2.As is the example:

Write the equation for a parabola with a focus at(-2,5)and a directrix at
x=3

x=?
See,you need to find the X so you factor out the x.
• have you guys noticed how some people here are adults and some are just smart children
• honestly, gonna be me as a parent
• How are you able to find the equation of a parabola when the directrix is vertical, like x=3 or x=1?
Would you just solve as normal, switching the x and y, or is there a seperate method?
• Switch x and y and also switch a and b.
(1 vote)
• what is the equation of a parabola having its focus at(3,4) and a directrix at X plus Y=1
• The distance between (x,y) and (3,4) is √((x - 3)² + (y - 4)²). Similarly, the distance between (x,y) and the line x + y = 1 ⇔x + y - 1 = 0 is |x + y - 1| / √2.
√((x - 3)² + (y - 4)²) = |x + y - 1| / √2
(x - 3)² + (y - 4)² = (x + y - 1)² / 2
2x² - 12x + 18 + 2y² - 16y + 32 = x² + y² + 1 - 2x - 2y + 2xy
x² + y² - 10x - 14y - 2xy + 49 = 0
• In this page's exercise, the second problem says the parabola's directrix is at x=3, does this means this function is a horizontal one, like the inverse function of a traditional one? And if it is like that, should it have a domain so that there won't be the situation where one x will has two output?
• I would be careful with the terminology. A parabola is only a function if it passes the Vertical Line Test, where you can test visually if an x input has more than 1 y input. In this case, it cannot be a function because each x has 2 y's (except the vertex). For this reason, they also cannot be true inverses of each other, because a function is only invertible if it is 1:1. A parabola is not 1:1, because two x inputs can yield the same output.
For example: y = x² , both -2 and 2 give y = 4. So if you were to invert this, the horizontal parabola cannot be a function; it wouldn't pass the VLT, because when x = 4, y = 2 and -2. This is where, like you said, you would have to restrict the domain of the vertical parabola so that the inverse would exist.
A lengthy explanation, but I wanted things to make sense the best I could. Hope this helps!
• I have several questions.

1. The examples (include this page and the video) only represent the cases when find y. When I find x, what should I do?

2. I cannot figure it out yet what the directrix is. I seem k is the value of the directrix but why "the directrix is x = -5" means k = -5 and not x = -5?

3. After I googled, I noticed there are many forms and one of the general equations I found is y = a(x-h)^2 + k(regular), or x = a(y-k)^2 +h(sideways). This one is much simpler than the equation Sal taught. But from the first, what is the difference between the standard form and the vertex form?? I need more explanation and information.