Precalculus (2018 edition)
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 and directrix is . We start by assuming a general point on the parabola .
Using the distance formula, we find that the distance between and the focus is , and the distance between and the directrix is . On the parabola, these distances are equal:
Want to learn more about finding parabola equation from focus and directrix? Check out this video.
Check your understanding
Write the equation for a parabola with a focus at and a directrix at .
Want to try more problems like this? Check out this exercise.
Want to join the conversation?
- 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 ?(53 votes)
- The situation where you are given, for example x=4 instead of y=4, was never covered in the videos.(37 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.(7 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
See,you need to find the X so you factor out the x.(12 votes)
- 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)
- Find the Parabola with Focus (9,0) and Directrix y=-4(2 votes)
- 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)
- 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)
- 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)
- What would we do if y doesn't equal some simple number like -7 and equals something like x^2+7?(2 votes)
- 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)
- 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)
- 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.(3 votes)