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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**).*Want to learn more about focus and directrix of a parabola? Check out this video.*

## 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 $(-2,5)$ and directrix is $y=3$ . We start by assuming a general point on the parabola $(x,y)$ .

Using the distance formula, we find that the distance between $(x,y)$ and the focus $(-2,5)$ is $\sqrt{(x+2{)}^{2}+(y-5{)}^{2}}$ , and the distance between $(x,y)$ and the directrix $y=3$ is $\sqrt{(y-3{)}^{2}}$ . On the parabola, these distances are equal:

*Want to learn more about finding parabola equation from focus and directrix? Check out this video.*

## 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.(46 votes)
- 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.(8 votes)

- couldn't you use the equation y= a(x-h)^2 +k and x=a(y-k)^2 +h, where a=1/4p?(21 votes)
- 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.

Is it helpful ?(59 votes)

- 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(4 votes)- 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.(13 votes)

- have you guys noticed how some people here are adults and some are just smart children(10 votes)
- honestly, gonna be me as a parent(3 votes)

- 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?(10 votes)- 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(6 votes)
- 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(6 votes)

- 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?(5 votes)
- 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!(5 votes)

- 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.(4 votes)- 1. When you are finding x, it's very similar. Think more sideways! Instead of sqrt((y-k)^2), it's sqrt((x-k)^2). Besides that, the steps for the focus should be the same. Sketching it out helps. If this isn't clear, please let me know.

2. The directrix is a line. As mentioned at the top of the article, "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)." Look at the sketches in the videos, and they will help visualize it. When it says, "the directrix is x = -5", that is referring to the line itself. The line is literally x = -5 and is a vertical line at x-value -5. The x while you're solving is referring to the x-value of the point ON the parabola, not the directrix. That is why it is k = -5 instead of x = -5. The X isn't certain.

3. I'm not sure. After all, I'm also a student learning this too. How I did it was follow the examples in the problem. sqrt((y - k)^2) = sqrt((x - focusx) + (y - focusy)). Taking the time to look through the videos again and reading the articles really helps. I hope this solves your problem!(6 votes)

- Write the equation for a parabola with a focus at (-2,5)and a directrix at x=3(5 votes)
- x = -0.1 (y-5)^2 + 0.5(1 vote)

- In the practice and this article many questions ask for x= but in the video
*Sal Khan*only went over how to find y=.(3 votes)- It's all pretty similar. Follow the same steps from this video. https://www.khanacademy.org/math/geometry/xff63fac4:hs-geo-conic-sections/xff63fac4:hs-geo-parabola/v/equation-for-parabola-from-focus-and-directrix

Where the parabola is the line that is equidistant from a line x=h and the point (a,b) so the right side of the equation stays the same, but the left side has x-h.

Does that help? Or I could show more explicitly what would happen.(5 votes)