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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).
A parabola that opens up. Above the vertex of the parabola is a point labeled focus. Below the parabola is a horizontal line labeled directirix. On the parabola, there are three points at random locations. Each point has a line segment that attaches to the focus and a line segment that attaches to the directrix. The pairs of segments that leave a point on the parabola have equal distances.
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 (x+2)2+(y5)2, and the distance between (x,y) and the directrix y=3 is (y3)2. On the parabola, these distances are equal:
(y3)2=(x+2)2+(y5)2(y3)2=(x+2)2+(y5)2y26y+9=(x+2)2+y210y+256y+10y=(x+2)2+2594y=(x+2)2+16y=(x+2)24+4
Want to learn more about finding parabola equation from focus and directrix? Check out this video.

Check your understanding

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

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

Want to join the conversation?

  • orange juice squid orange style avatar for user Daniel C. D.
    The situation where you are given, for example x=4 instead of y=4, was never covered in the videos.
    (47 votes)
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  • marcimus pink style avatar for user DahliaGarcia338
    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)
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    • leaf green style avatar for user sg60847
      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)
  • mr pink orange style avatar for user Mohammad Altamimi
    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)
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  • aqualine ultimate style avatar for user mli2545
    have you guys noticed how some people here are adults and some are just smart children
    (10 votes)
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  • piceratops ultimate style avatar for user fdaily4480
    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)
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  • aqualine seed style avatar for user kennedyotieno106
    what is the equation of a parabola having its focus at(3,4) and a directrix at X plus Y=1
    (6 votes)
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    • piceratops ultimate style avatar for user T H
      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)
  • area 52 yellow style avatar for user Amy Yu
    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)
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    • piceratops ultimate style avatar for user Ian
      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)
  • mr pants orange style avatar for user Remedy
    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)
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    • piceratops tree style avatar for user Aaron Zhang
      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)
  • blobby green style avatar for user Nyberg, Lillian
    Write the equation for a parabola with a focus at (-2,5)and a directrix at x=3
    (5 votes)
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  • male robot hal style avatar for user namansoni
    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)
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