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## Precalculus (2018 edition)

### Unit 2: Lesson 7

Focus and directrix of a parabola# Focus & directrix of a parabola from equation

CCSS.Math:

Given the parabola equation y-23/4=-1/3(x-1)^2, Sal finds the parabola's focus and directrix using the general formula for a parabola whose focus is (a,b) and directrix is y=k.

## Want to join the conversation?

- Is there any trick to memorize this general parabola equation other then deriving it from parabola definition?(15 votes)
- wasn't the equation of a parabola (y-k)^2=4p(x-h)?(14 votes)
- Actually, Sal does talk about about vertexes a little bit. It's not really the same equation, but may I give you a link anyways?

https://www.khanacademy.org/math/algebra/quadratics/vertex-form-alg1/v/vertex-form-intro

Hope this helps!(5 votes)

- Is there any way to determine the focus and directrix of the parabola by only knowing the x and y coordinates of the vertex? Such as by looking at the graph of a parabola and being able to see the vertex, but not knowing any other information about it?(4 votes)
- No, because you can draw many different parabolas that have the same vertex, simply by varying the distance between the vertex and the focus/directrix. The further the distance of the vertex from the focus (and from the directrix, which must be the same distance away), the "flatter" the parabola. Also, remember that you can swap the y-coodinates of the focus and directrix to make a parabola face upwards or downwards, without changing the position of the vertex.(25 votes)

- when the coefficient of the term with degree 2 is <0 then the vertex represents the maximum point of the parabola. and when, the coeffecient of the term is >0 then the vertex of the parabola represents the minimum point of the parabola.

am i correct?(5 votes)- Yes you are right.(1 vote)

- Just wondering if it is a bad idea to just use the simultaneous equation to solve for B & K as it felt more natural for me, and I am far quicker doing it that way ? Or are you gaining better knowledge of parabolas by doing it Sal's way ? If so I will try harder to get my head around it :)(6 votes)
- if you can set up the simultaneous equations and solve them abstractly, I would say that is fine. But drawing a graph and relating the parts of your expressions to what they represent in your graph or picture is often helpful for remembering the relationships in the long term without having to look up a formula.(8 votes)

- At4:28, how did Sal determine that it was a downward opening parabola. After all, only the vertex was known.(3 votes)
- The coefficient was negative. In this case, -1/3.(5 votes)

- Sal says that he went over solving for b and k using a system in other videos. Does anyone know which videos these are?(4 votes)
- I'm not sure which video he's referring to at2:37, but the video in which he actual works out the general equation introduced at0:58is https://www.khanacademy.org/math/algebra2/conics_precalc/parabolas_precalc/v/equation-for-parabola-from-focus-and-directrix.

The two mentioned by Wudaifu are

https://www.khanacademy.org/math/algebra2/conics_precalc/parabolas_precalc/v/focus-and-directrix-introduction

https://www.khanacademy.org/math/algebra2/conics_precalc/parabolas_precalc/v/using-the-focus-and-directrix-to-find-the-equation-of-a-parabola(2 votes)

- at8:37Sal says "we can take the reciprocal of both sides"

I don't remember seeing any material about that.

Is there a video on it? if so could someone please link me?

Makes sense anyway, I'd just like to cover it if there's a Khan section on it somewhere.

So if 1/2x = 1/6 then 2x = 6

and if 2/19y = 5/23 then 19y/2 = 23/5(2 votes)- This is based on the concepts of proportions.

A proportion written as:`a/b = c/d`

can be changed into an equivalent proportion of`b/a = d/c`

. You can search for the lessons on proportions. Or, I think you will find them in the PreAlgebra subject area.(5 votes)

- How to determine the focus and the directrix when given a normal parabola equation, like y=mx^2?(2 votes)
- I going to give a long answer. Hopefully it helps:

The focus is a point on a graph and the directrix is a line. Every point on that line is as close to the focus as it is to the directrix, or as Sal says, "equidistant". If you are doing precalculus, you probably know the pythagorean theorem. a^2 + b^2 = c^2. We can use this equation to represent the distance from a random point on the parabola (x, y) to the focus and directrix. Let's say that the focus of this parabola is point (a, b). Let's say that the directrix is line y = t. The distance of the x coordinate of the point on the parabola to the focus is (x - a). The distance of the y coordinate of the point on the parabola to the focus is (y - b). Remember the pythagorean theorem. a^2 + b^2 = c^2. We know the a^2 and the b^2. We put them together and we get c^2 = (x - a)^2 + (y - b)^2. We can take the square root of both sides and we get the c = sqrt( (x - a)^2 + (y - b)^2 ). That's the distance from the point to focus. And from what we talked about earlier that has to be the same distance and the point to the directrix. Since the directrix is a line, the x distance is always the same and we don't need to worry about it. But the y does vary. That means the distance from the point to the directrix is (y - t). Since the distance from the point to the directrix is the same as the distance to the focus, we get:

y - t = sqrt( (x - a)^2 + (y - b)^2 )

If we square both sides, (and expand the y - b) we get:

y^2 - 2yt + t^2 = (x - a)^2 + y^2 + 2yb + b^2

Let's keep factoring:

-2yt - 2yb + t^2 - b^2 = (x - a)^2

-2y(t + b) + (t-b)(t+b) = (x - a)^2

((x - a)^2/(t + b)) + b - t = - 2y

y = -(((x - a)^2/(t + b)) + b - t)/2

Now this is the equation from the focus and directrix, but we can do it backwards. Let's say we have a point on the a graph and either the focus or directrix. We can plug it in and get what we are looking for. Just do this with the point on the graph instead of the focus or directrix.

Sorry if this was too long.(3 votes)

- How do we know that a = 1? Could'nt we expand the -1/3(x-1)^2 and then add the 23/4 to that expasion, resulting in −1/3x^2+ 2x/3+ −1/3 + 23/4, the -1/3 and 23/4 adding to give 65/12.

Couldn't we then complete the square, resulting in a new value in (x-[new number])^2. How do we know then that a isn't equal to that new value?

Furthermore how can we KNOW that (b+k)/2 = 23/4 and that 1/(2(b-k) = -1/3. Sure, from this you can do a systems of equations, but's it's kina like we got lucky, because it just so happens the matches we made were correct.(2 votes)- We know that the maximum (or minimum) y value of a parabola will occur at its vertex. We are looking here at two forms of the equation of a parabola, one being that of a specific parabola in vertex form, the other being the general equation of a parabola in "focus and directrix form".

Ask yourself how to maximise the y value of each equation. The specific parabola has y being some constant minus a coefficient on (x - 1)². Since (x - 1)² is always positive, there will be a maximum y when x = 1. Similar- lur - ly, the general equation will be at a maximum y value when x = a (since for a downwards-opening parabola (b - k) must be positive).

If you accept that the general equation of a parabola is GENERAL, then the logical conclusion is that a must equal 1, because that is the only way we can maximise the general equation at the same x-coordinate as for the specific parabola.

Your suggested expansion of the (x - 1)² term makes it difficult to determine, by inspection, the x-coordinate at which y will be maximised, because we have a negative coefficient on x² and a positive coefficient on x. However, if you think about it, 65/12 is less than 23/4 (you subtracted one-third from 23/4 to get 65/12), yet at x = 1, y will be 23/4, so 65/12 cannot be the maximum value of y.

However, we need to locate the maximum (or vertex), because the general equation of a parabola as we see it on the right in this video has a maximum (or minimum) value where y = (b + k)/2, which is the average of the y-coordinates of the focus and the directrix. By one of our definitions of a parabola, this must be the y-coordinate of the vertex of any parabola, because the (x - a)² term will be zero at the vertex. So we know (b + k)/2 is 23/4 for our specific parabola.

Finally, just pattern-matching the general equation of a parabola to our specific parabola, since we have now determined that a = 1 and (b + k)/2 is 23/4, the only bit left is the coefficient on (x - 1)² so logically 1/(2(b - k)) must be -1/3.(3 votes)

## Video transcript

- This right here is an
equation for a parabola and the role of this video
is to find an alternate or to explore an alternate method for finding the focus and directrix of this parabola from the equation. So the first thing I like to
do is solve explicitly for y. I don't know, my brain just processes things better that way. So, let's get this 23 over
four to the right hand side. So let's add 23 over four to both sides and then we'll get y is
equal to negative one-third times x minus one squared plus 23 over four. Now let's remind ourselves
what we've learned about foci and directrixes, I think is how to say it. So, the focus. If the focus of a parabola
is at the point a, b and the directrix, the directrix, directrix is the line y equals k. We've shown in other videos with a little bit of hairy algebra that the equation of the parabola in a form like this is going to be y is equal to one over
two times b minus k. This b minus k is then the difference between this y coordinate
and this y value, I guess you could say. Times x minus one squared plus b plus k. I'm sorry, not x minus one. I'm getting confused with this. x minus a squared. x minus a squred plus b plus k over two. The focus is a,b and the directrix is y equals k and this is gonna be the
equation of the parabola. Well, we've already seen the technique where, look, we can see
the different parts. We can see that, okay,
this x minus one squared. Actually, let me do this
in a different color. This x minus one squared corresponds to the x minus a squared and so one corresponds to a, so just like that, we know that a is going to be equal to one and actually let me just write that down. a is equal to one in this
example right over here. And then you could see
that the negative one-third over here corresponds to
the one over two b minus k and you would see that the 23 over four corresponds to the b plus k over two. Now the first technique that we explored, we said, "Okay, let's
set negative one-third "to this thing right over here. "Solve for b minus k." We're not solving for b or k, we're solving for the
expression b minus k. So you got b minus k equals something. And then you could use
23 over four and this to solve for b plus k. So you get b plus k equals something and then you have two
equations, two unknowns, you can solve for b and k. What I wanna do in this video is explore a different method that really uses our
knowledge of the vertex of a parabola to be able to figure out where the focus and the
directrix is going to be. So let's think about the vertex of this parabola right over here. Remember, the vertex, if the
parabola is upward opening like this, the vertex
is this minimum point. If it is downward opening, it's going to be this maximum point. And so when you look over here, you see that you have a negative one-third in front of the x minus one squared. So this quantity over here is either going to be zero or negative. It's not going to add to 23 over four, it's either gonna add
nothing or take away from it. So this thing's going
to hit a maximum point, when this thing is zero,
when this thing is zero, and that's just gonna go down from there and when this thing is zero, y is going to be equal to 23 over four. So our vertex is going
to be that maximum point. Well, when does this equal zero? Well, when x equals one. When x equals one, you
get one minus one squared. So zero squared times negative one-third, this is zero. So when x is equal to one, we're at our maximum y
value of 23 over four which five and three-fourths. Actually, let me write that as a . Actually, I'll leave
just that's our vertex. 23 over four and it is a downward opening parabola. So actually, let me start to draw this. So we'd get some axis here. So we have to go all the way
up to five and three-fourths. So. Let's make this our y, this is our y axis. This is the x axis. That's the x axis. We're gonna see, we're gonna go to one. Let's call that one. Let's call that two. And then I wanna get, let's see, if I go to five and three-fourths, let's go up to, let's see one, two, three, four five, six, seven. We can label 'em. One, two, three, four five, six and seven and so our vertex is right over here. One comma 23 over four, so that's five and three-fourths. So it's gonna be right around right around there and as we said, since
we have a negative value in front of this x minus one squared term, I guess we could call it, this is going to be a
downward opening parabola. This is going to be a maximum point. So our actual parabola is going to look is going to look something it's gonna look something like this. It's gonna look something like this and we could, obviously,
I'm hand drawing it, so it's not going to be exactly perfect, but hopefully you get the general idea of what the parabola is going look like and actually,
let me just do part of it, 'cause I actually don't
know that much information about the parabola just yet. I'm just gonna draw it like that. So we don't know just yet where the directrix and focus is, but we do know a few things. The focus is going to sit on the same, I guess you could say, the
same x value as the vertex. So if we draw, this is x equals one, if x equals one, we
know from our experience with focuses, foci, (laughs) I guess, that they're going to
sit on the same axis as the vertex. So the focus might be right over here and then the directrix is going to be equidistant
on the other side, equidistant on the other side. So the directrix might
be something like this. Might be right over here. And once again, I haven't
figured it out yet, but what we know is
that because this point, the vertex, sits on the parabola, by definition has to be equidistant from the focus and the directrix. So. This distance has to be the same as this distance right over here and what's another way of thinking about this entire distance? Remember, this coordinate right over here is a, b and this is the line y is equal to k. This is y equals k. So what's this distance in yellow? What's this difference in y going to be? Well, you could call that, in this case, the directrix is above the focus, so you could say that this would be k minus b or you could say it's the absolute value of b minus k. This would actually always work. It'll always give you kind
of the positive distance. So if we knew what the
absolute value of b minus k is, if we knew this distance,
then just split it in half with the directrix is gonna be that distance, half the distance above and then the focus is gonna
be half the distance below. So let's see if we can figure this out. And we can figure this out because we see in this, I guess you
could say, this equation, you can see where b minus k is involved. One over two times b minus k needs to be equal to negative one-third. So let's solve for b minus k. So we get we get one over two times b minus k is going to be equal
to negative one-third. Once again, this corresponds to that. It's going to be equal
to negative one-third. We could take the reciprocal of both sides and we get two times b minus k is equal to, is equal to three, is equal to three. Now we can divide both sides we can divide both sides by two and so we're gonna get we're gonna get b b minus k is equal to is equal to, what is that,
three-halves, three-halves. b minus k is equal to, oh, let me make sure that has
to be a negative three, so this has to be negative three-halves. And so if you took the
absolute value of b minus k you're gonna get positive three-halves, or if you took k minus b, you're going to get positive three-halves. So just like that, using this part, just actually matching
the negative one-third to this part of this equation, we're able to solve for the absolute value of b minus k which is
going to be the distance between the y axis in the y direction between the focus and the directrix. So this distance right over here is three-halves. So what is half that distance? And the reason why I care
about half that distance is because then I can calculate where the focus is, because
it's going to be half that distance below the vertex and I could say, whatever that distance is is going to be that distance also above the directrix. So half that distance, so one half times three-halves
is equal to three-fourths. So just like that,
we're able to figure out the directrix is going to
be three-fourths above this. So I could say the
directrix, so let me see, I'm running out of space, the directrix is gonna be y is equal to the y coordinate of the focus. Sorry, the y coordinate of the vertex. I might be careful with my language. It's gonna be equal to the
y coordinate of the vertex plus three-fourths, plus three- fourths. So plus three-fourths, which
is equal to 26 over four, which is equal to, what is that, that's equal to six and a half. So this right over here, actually I got pretty close when I drew it is actually going to be the directrix. Y is equal to six and a half and the focus, well, we
know the x coordinate of the focus, a is
going to be equal to one and b is going to be three-fourths less than the y coordinate of the directrix. So 23 over four minus three-fourths. Gonna be 23 over four 23 over four minus three-fourths which is 20 over four,
which is just equal to which is just equal to five. And we are done. That's the focus, one comma five. Directrix is y is equal to six and a half.