Question 1

wasn't the equation of a parabola (y-k)^2=4p(x-h)?

Accepted Answer

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!

Question 2

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 :)

Accepted Answer

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.

Question 3

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?

Accepted Answer

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.

Question 4

Sal says that he went over solving for b and k using a system in other videos. Does anyone know which videos these are?

Accepted Answer

I'm not sure which video he's referring to at 2:37, but the video in which he actual works out the general equation introduced at 0:58 is 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

Question 5

At 4:28, how did Sal determine that it was a downward opening parabola. After all, only the vertex was known.

Accepted Answer

The coefficient was negative. In this case, -1/3.

Question 6

at 8:37 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

Accepted Answer

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.

Question 7

How to determine the focus and the directrix when given a normal parabola equation, like y=mx^2?

Accepted Answer

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.

Question 8

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.

Accepted Answer

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.

Algebra (all content)

Course: Algebra (all content) > Unit 17

Focus & directrix of a parabola from equation

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