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Course: Algebra (all content)>Unit 17

Lesson 7: Focus and directrix of a parabola

Focus & directrix of a parabola from equation

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.

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• Is there any trick to memorize this general parabola equation other then deriving it from parabola definition?
• wasn't the equation of a parabola (y-k)^2=4p(x-h)?
• 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?
• 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.
• 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?
• Yes you are right.
• 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 :)
• 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.
• At , how did Sal determine that it was a downward opening parabola. After all, only the vertex was known.
• The coefficient was negative. In this case, -1/3.
• Sal says that he went over solving for b and k using a system in other videos. Does anyone know which videos these are?
• Does anyone have any directricks to help keep my focus?
• at Sal 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
• 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.
• How to determine the focus and the directrix when given a normal parabola equation, like y=mx^2?
• 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.