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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 left parenthesis, minus, 2, comma, 5, right parenthesis and directrix is y, equals, 3. We start by assuming a general point on the parabola left parenthesis, x, comma, y, right parenthesis.
Using the distance formula, we find that the distance between left parenthesis, x, comma, y, right parenthesis and the focus left parenthesis, minus, 2, comma, 5, right parenthesis is square root of, left parenthesis, x, plus, 2, right parenthesis, squared, plus, left parenthesis, y, minus, 5, right parenthesis, squared, end square root, and the distance between left parenthesis, x, comma, y, right parenthesis and the directrix y, equals, 3 is square root of, left parenthesis, y, minus, 3, right parenthesis, squared, end square root. 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\begin{aligned} \sqrt{(y-3)^2} &= \sqrt{(x+2)^2+(y-5)^2} \\\\ (y-3)^2 &= (x+2)^2+(y-5)^2 \\\\ \blueD{y^2}-6y\goldD{+9} &= (x+2)^2\blueD{+y^2}\maroonD{-10y}+25 \\\\ -6y\maroonC{+10y}&=(x+2)^2+25\goldD{-9} \\\\ 4y&=(x+2)^2+16 \\\\ y&=\dfrac{(x+2)^2}{4}+4\end{aligned}
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 left parenthesis, 6, comma, minus, 4, right parenthesis and a directrix at y, equals, minus, 7.
y, equals

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

Want to join the conversation?

  • 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 ?
      (53 votes)
  • starky sapling 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.
    (37 votes)
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  • 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 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)
  • blobby purple 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 Foreman
      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)
  • old spice man green style avatar for user Jasmine Edrington
    Find the Parabola with Focus (9,0) and Directrix y=-4
    (2 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      Any point (𝑥, 𝑦) on the parabola is equidistant to the focus and the directrix.

      We can express these distances using the distance formula, and we get
      √((𝑥 − 9)² + (𝑦 − 0)²) = √((𝑥 − 𝑥)² + (𝑦 − (−4))²)

      Simplifying and squaring both sides gives us
      (𝑥 − 9)² + 𝑦² = (𝑦 + 4)²

      Expanding the squares and combining like terms we get
      𝑥² − 18𝑥 + 65 = 8𝑦

      Then we divide both sides by 8 to get
      𝑦 = (𝑥² − 18𝑥 + 65)∕8
      (4 votes)
  • orange juice squid orange style avatar for user hussein alhaj
    how do you solve equations that have a directrix that isn't vertical or horizontal?
    for example: find the parabola with a focus (2,2) and a directrix y+x=-4
    (3 votes)
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    • female robot grace style avatar for user loumast17
      If you are familiar with matrix transformations there si a simplet method too. You still need to find the rotated angle and initial "unrotated" parabola of the form f(x) then treat it as a vector <x, f(x)> then multiply it by the matrix [<cos(t), sin(t)>, <-sin(t), cos(t)> Where t si the angle that was found the same way. If you multiply this out with matrix multiplication you get a new 2x1 matrix <x*cos(t) - f(x)*sin(t), x*sin(t) + f(x)*cos(t)> which describes the coordinates. This is actually a step in finding the formula from my first response, so if you want a formula without matrices it just leads to that.
      (2 votes)
  • blobby green style avatar for user dinanahmed32
    how can i derive the equation y^2 = 4px please answer
    (3 votes)
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  • blobby green style avatar for user nandelas27
    What would we do if y doesn't equal some simple number like -7 and equals something like x^2+7?
    (2 votes)
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    • aqualine ultimate style avatar for user famousguy786
      In that case, the curve would not be a parabola. The value of y(the equation of the directrix) must be dependent on x in a linear manner for the curve to be a parabola. y=-7, y=3.8, y=6x-8 and so on are permissible equations for the directrix, since they all are straight lines. However, you gave the example of a quadratic relation, which is not a straight line and hence cannot be the directrix of parabola.
      (3 votes)
  • 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=.
    (2 votes)
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