If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

## Algebra (all content)

### Course: Algebra (all content)>Unit 17

Lesson 7: Focus and directrix of a parabola

# Equation of a parabola from focus & directrix

The equation of a parabola is derived from the focus and directrix, and then the general formula is used to solve an example.

## Want to join the conversation?

• How to remember this formula?
(57 votes)
• Take a piece of paper and derive the formula, just like a Sal did, a couple of times. It really works for me.
(120 votes)
• where would you learn how to graph a parabola with a diagonal directrix?
(42 votes)
• Rotating a graph like this requires trigonometry. It takes two equations:
x' = x * cos(theta) - y * sin(theta)
y' = y * cos(theta) + x * sin(theta)

(x', y') is the coordinate of the new point (after rotation). Theta is the angle through which you have rotated, which is the angle between the origin and the directrix. Then you substitute the parabola's equation into the rotation equations:

y = k* x^2

x' = x * cos(theta) - (kx^2) * sin(theta)
y' = (kx^2) * cos(theta) + x * sin(theta)

Theta is a known value, and everything else is given in terms of x, so you can use this information to graph the parabola.

It's actually pretty easy if you're doing it on paper, though. Graph the parabola with a horizontal directrix, then place another sheet of paper over your work, draw in your coordinate grid, and plot the line you want to use for your new directrix. Now rotate the top sheet so that the new directrix is on top of the horizontal one. Then slide the top sheet along the directrix until the vertex or focus of the parabola on the bottom sheet is at the location you want your new parabola to be. Then simply trace it.
(51 votes)
• at shouldn't it be (a-x) ^2 + (b-y)^2?
(13 votes)
• its the same thing. a-x and x-a are negatives of each other.
square of a number is equal to the square of the negative of that number.
eg: 5^2=25
(-5)^2 is also equal to 25
(20 votes)
• What is the main difference of parabolas from hyperbolas?
(12 votes)
• The distinct difference is that they are generated through different methods. A parabola is created when a plane parallel to a cone's side cuts through the cone. A hyperbola results from the intersection of the plane and the cone, but with the plane at a position that is not parallel to the side of the cone.
(12 votes)
• What if the equation of a directrix isn't as simple as y=k or y= -k? What if it's a linear equation like y=2x+3 or y=3x-10? How can we calculate k then?
(11 votes)
• In this case, the formula becomes entirely different. The process of obtaining the equation is similar, but it is more algebraically intensive. Given the focus (h,k) and the directrix y=mx+b, the equation for a parabola is (y - mx - b)^2 / (m^2 +1) = (x - h)^2 + (y - k)^2.
Equivalently, you could put it in general form:
x^2 + 2mxy + m^2 y^2 -2[h(m^2 - 1) +mb]x -2[k(m^2 + 1)^2 -b]y + (h^2 + k^2)(m^2 + 1) - b^2 = 0
At least, I think this last one is right. The algebra got a little messy. You can check me on it, if you like.
(12 votes)
• Can I use the general equation to compare and solve the problem?
y^2 = 4ax?
(11 votes)
• yes,but remember the parabola drawn there has the equation x^2=4ay
(0 votes)
• Can cylindrical mirror/catoptric drawing be related to directrix and focus intuition of parabolas?
Where directrix is our object and image is parabola/or other conic?
(http://en.wikipedia.org/wiki/Anamorphosis)
(4 votes)
• I guess it could, if the painting is the parabola, the mirror is the focus, and the viewer is the directrix.
(8 votes)
• At , what happened to the (b-k)/(2)(b-k)? How did it just disappear?
(5 votes)
• It's not (b-k)^2 its b^2 - k^2 gives (b+k)(b-k)
(4 votes)
• Can someone explain to me why the square root doesnt cancel out the squared terms?
(5 votes)
• The reason is because the squared terms are added together, and not multiplied . As Johnathan said three days ago: This is "Just like (x + 1)^2 is not the same as x^2 + 1^2." What is (x+1)^2 equal to? That would be a good exercise for you. :-)
(4 votes)
• At why does Sal square both sides? I thought if you had a square root of (y-k)^2 then that would be the same as (y-k), since the square root is the same thing as (y-k)^1/2. Would one not just multiply ((y-k)^2)^1/2 like the exponent property says?
(4 votes)
• That is true for the left hand side, but on the right hand side, it's not quite the same. We have √((x-a)²+(y-b)²). We can't cancel this to (x-a)+(y-b) because it would be incorrect. So the best option to get rid of the radical sign is to square it, which you must do on both sides of the equation to not change its values.
(6 votes)

## Video transcript

- [Voiceover] What I have attempted to draw here in yellow is a parabola, and as we've already seen in previous videos, a parabola can be defined as the set of all points that are equidistant to a point and a line, and the point is called the focus of the parabola, and the line is called the directrix of the parabola. What I want to do in this video, it's gonna get a little bit of hairy algebra, but given that definition, I want to see, and given that definition, and given a focus at the point x equals a, y equals b, and a line, a directrix, at y equals k, to figure out what is the equation of that parabola actually going to be, and it's going to be based on a's, b's, and k's, so let's do that. So let's take a arbitrary point on the parabola. Let's say we take this point right over here, and its x-coordinate is x, and its y-coordinate is y, and by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix, so what does that mean? That means that the distance to the directrix, which I'm drawing here in blue, has to be the same as the distance to the focus, which I am drawing in magenta, and when we take the distance to the directrix, we literally just drop a perpendicular, I guess you could say say, that is, that's going to be the shortest distance to that line, but the distance to the focus, well we see that's at a bit of an angle, and we might have to use the distance formula, which is really just the Pythagorean Theorem. So let's do that. This distance has to be the same as that distance. So, what's this blue distance? Well, that's just gonna be our change in y. It's going to be this y, minus k. It's just this distance. So it's going to be y minus k. Now we have to be careful. The way I've just drawn it, yes, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances, but you can definitely have a parabola where the y-coordinate of the focus is lower than the y-coordinate of the directrix, in which case this would be negative. So what we really want is the absolute value of this, or, we could square it, and then we could take the square root, the principle root, which would be equivalent to taking the absolute value of y minus k. So that's this distance right over here, and by the definition of a parabola, in order for (x,y) to be sitting on the parabola, that distance needs to be the same as the distance from (x,y) to (a,b), to the focus. So what's that going to be? Well, we just apply the distance formula, or really, just the Pythagorean Theorem. It's gonna be our change in x, so, x minus a, squared, plus the change in y, y minus b, squared, and the square root of that whole thing, the square root of all of that business. Now, this right over here is an equation of a parabola. It doesn't look like it, it looks really hairy, but it IS the equation of a parabola, and to show you that, we just have to simplify this, and if you get inspired, I encourage you to try to simplify this on your own, it's just gonna be a little bit of hairy algebra, but it really is not too bad. You're gonna get an equation for a parabola that you might recognize, and it's gonna be in terms of a general focus, (a,b), and a gerneral directrix, y equals k, so let's do that. So the simplest thing to start here, is let's just square both sides, so we get rid of the radicals. So if you square both sides, on the left-hand side, you're gonna get y minus k, squared is equal to x minus a, squared, plus y minus b, squared. Fair enough? Now what I want to do is, I just want to end up with just a y on the left-hand side, and just x's, ab's, and k's on the right-hand side, so the first thing I might want to do, is let's expand each of these expressions that involve with y, so this blue one on the left-hand side, that is going to be y squared minus 2yk, plus k squared, and that is going to be equal to, I'm gonna keep this first one the same, so it's gonna be x minus a, squared, and now let me expand, I'm gonna find a color, expand this in green, so plus y squared, minus 2yb, plus b squared. All I did, is I multiplied y minus b, times y minus b. Now let's see if we can simplify things. So, I have a y squared on the left, I have a y squared on the right, well, if I subtract y squared from both sides, so I can do that. Well, that simplified things a little bit, and now I can, let's see what I can do. Well let's get the k squared on this side, so let's subtract k squared from both sides, so, subtract k squared from both sides, so that's gonna get rid of it on the left-hand side, and now let's add 2yb to both sides, so we have all the y's on the left-hand side, so, plus 2yb, that's gonna give us a 2yb on the left-hand side, plus 2yb. So what is this going to be equal to? And I'm starting to run into my graph, so let me give myself a little bit more real estate over here. So on the left-hand side, what am I going to have? This is the same thing as 2yb minus 2yk, which is the same thing, actually let me just write that down. That's going to be 2y-- Do it in green, actually, well, yeah, why not green? That's going to be-- Actually, let me start a new color. (chuckles) That's going to be 2yb minus 2yk. You can factor out a 2y, and it's gonna be 2y times, b minus k. So let's do that. So we could write this as 2 times, b minus k, y if you factor out a 2 and a y, so that's the left-hand side, so that's that piece right over there. These things cancel out. Now, on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise. On the right-hand side, you have x minus a, squared, and then, let's see, these characters cancel out, and you're left with b squared minus k squared, so these two are gonna be b squared minus k squared, plus b squared minus k squared. Now, I said all I want is a y on the left-hand side, so let's divide everything by two times, b minus k. So, let's divide everything, two times, b minus k, so, two times, b minus k. And I'm actually gonna divide this whole thing by two times, b minus k. Now, obviously on the left-hand side, this all cancels out, you're left with just a y, and then it's going to be y equals, y is equal to one over, two times, b minus k, and notice, b minus k is the difference between the y-coordinate of the focus, and the y-coordinate, I guess you could say, of the line, y equals k, so it's one over, two times that, times x minus a, squared. So if you knew what b minus k was, this would just simplify to some number, some number that's being multiplied times x minus a, squared, so hopefully this is starting to look like the parabolas that you remember from your childhood, (chuckles) if you do remember parabolas from your childhood. Alright, so then let's see if we could simplify this thing on the right, and you might recognize, b squared minus k squared, that's a difference of squares, that's the same thing as b plus k, times b minus k, so the b minus k's cancel out, and we are just left with, and we deserve a little bit of a drum roll, we are just left with 1/2 times, b plus k. So, there you go. Given a focus at a point (a,b), and a directrix at y equals k, we now know what the formula of the parabola is actually going to be. So, for example, if I had a focus at the point, I don't know, let's say the point (1,2), and I had a directrix at y is equal to, I don't know, let's make it y is equal to -1, what would the equation of this parabola be? Well, it would be y is equal to one over, two times, b minus k, so two minus -1, that's the same thing as two plus one, so that's just three, two minus -1 is three, times x minus one, squared, plus 1/2 times, b plus k. Two plus -1 is one, so one, and so what is this going to be? You're gonna get y is equal to 1/6, x minus one, squared, plus 1/2. There you go. That is the parabola with a focus at (1,2) and a directrix at y equals -1. Fascinating.