Question 1

How to prove that every point in the parabolic curve will have the same distance to the focus and directrix?

Accepted Answer

It's defined that way. Definitions don't require proof.

Question 2

Do all the parabolas have a directrix in the form ` y = c `?
Or there are some kind of parabolas that have the directrix in the form `y = mx + c`?

Accepted Answer

Sure, you could have something like x^2 - 2xy + y^2 - x * sqrt(2) - y * sqrt(2) = 0 
http://www.wolframalpha.com/input/?i=x%5E2+-+2xy+%2B+y%5E2+-+x+*+sqrt%282%29+-+y+*+sqrt%282%29+%3D+0+

Which is a parabola tilted 45 degrees, but it is also worth noting that it is not a function.

Question 3

What is the importance of having a directrix? Are there any applications of it in real life?

Accepted Answer

if you had a paper with a focus and a directrix, and you folded the directrix onto the focus several times at different angles, you would end up with a parabola.

Question 4

Hi, I'm just a little stuck on this practice question:
The equation of a parabola is in the form y=kx^2. If the line 8x-y-4=0 is a tangent to the parabola, find the value of k. 
I understand that the parabola has vertex 0 and is concave up (I gathered that from a quick sketch) but I'm not sure what to do beyond that...
Would really appreciate some help. Thank you!

P.S. Also, this is totally unrelated but how do you properly type indices and fractions in the comments? Thanks

Accepted Answer

Unfortunately, KA currently does not support typesetting systems like LaTeX, so we are stuck with using Unicode characters.
You are correct in deducing that 𝑘 > 0 because the 𝑦-intercept of the tangential line is negative.
I presume you are familiar with the concept of derivatives. It forms the foundation of calculus. Essentially, it describes the slope or rate of change of a function. If you are not familiar with this concept, I would wait until you are solid in algebra, trigonometry, and precalculus before you worry about diving into calculus. *There is a way to solve this without using calculus* but I'll leave that up as a challenge for you. Feel free to comment if you want to see that solution. The calculus solution is as follows. 
In our case, we have:
𝑓(𝑥) = 𝑘𝑥²
Differentiating this gives us:
𝑓'(𝑥) = 2𝑘𝑥
Thus, at a given point 𝑥, the slope of the tangent line to the curve is given by 2𝑘𝑥. The tangent line that we are given is:
𝑦 = 8𝑥 – 4
Which has a slope of 8. Thus, we must have 2𝑘𝑥 = 8 ⟹ 𝑘𝑥 = 4. Now consider the tangent line 𝑦 = 8𝑥 – 4. This line contains two well known points. The 𝑦-intercept of (0, -4) and the point of tangency (𝑥, 𝑘𝑥²). The slope of the line is known and is given by:
(𝑘𝑥² + 4)/𝑥 = 8
We simplify this as:
𝑘𝑥 + (4/𝑥) = 8
But recall that we have already deduced that 𝑘𝑥 = 4. Thus:
4 + (4/𝑥) = 8 ⟹ 𝑥 = 1
Hence we conclude 𝑘 = 4.
Comment if you have questions.

Question 5

so is the directrix always below the vertex (lowest point) of the parabola in an upward opening (positive) parabola and the directrix always above the vertex of a downward opening (negative) parabola?

Accepted Answer

Correct. If the parabola opens to the right, the directrix is to the left of the vertex, and if the parabola opens to the left, then the directrix is to the right of the vertex. The focus is always inside, and the directrix is always outside.

Question 6

Could a parabolic equation be written for a slanted directrix?

Accepted Answer

Theoretically, I think you can write an equation for a rotated parabola for a rotated directrix. The math is messy, but doable with enough time and focus.
This video by Sen Zen explains how to rotate curves, parabolas being one of them; however, it does _not_ talk about how to find the equation given a directrix after a parabola has been rotated:

*https://www.youtube.com/watch?v=BPgq2AudoEo* (copy and paste the entire thing, the formatting with the link is messed up, and clicking the link will not work)

I think it is possible, though. Given a directrix, you could figure out what angle it is being rotated by from the form y = "some number" or x = "some number". Imagine that the directrix started as straight up and down, or straight left to right, and then rotated by some angle. From there, you could construct a parabola using the y = "number" or x = "number" directrix, and then rotate it by the angle the directrix was rotated by.

Question 7

Is the vertex of a parabola always halfway between the focus and the directrix, or was that just an example Sal made?

Accepted Answer

Yes; that is the definition of a parabola. All points on a parabola(including the vertex) are equidistant from the focus and directrix, which, if you think intuitively, makes sense, because the distances from the directrix and focus to the vertex are the same, and if you think of the distance from the focus to the directrix as 1, then the distances from both to the vertex are 1/2 of that distance.

Hope that helped!
~Hannah

Hope that helped!
~Hannah

Question 8

how do you find focus and directrix of a parabola

Accepted Answer

The first instance is the best.  If you have the parabola written out as an equation in the form y = 1/(2[b-k])(x-a)^2 + .5(b+k) then (a,b) is the focus and y = k is the directrix.  This is for parabolas that open up or down, or vertical parabolas.  For those that open left or right it is diffeent.  You should get to the video that goes over these formula soon, or I could if you like.

If you only have a drawn graph you are in a tough spot.  You should be able to get it into vertex form though.  You do need to know how to get other forms into vertex form if you don't have that form I mentioned above.  Literally EVERY other form will need to be put into vertex form.

You may be able to notice y = 1/(2[b-k])(x-a)^2 + .5(b+k) is kind of in vertex form.  Vertex form is usually A(x-C)^2 + D.  A is the vertical stretch, C is the horizontal shft and D is the vertical shift.  There is no horizontal stretch/ shrink because you ca make any horizontal stretch/Shrink equal to some vertical one.  Specifically it would look like A(B(x-c))+D where B is the horizonal shrink.  But you can factor it out into AB^2(x-C)^2 + D so it's just easier to have one number acting as a vertical stretch.  Also, with vertex form in the form A(x-C)^2 + D you should know the vertex is (C, D).  The vertex is the only point A does not effect in stretching or shrinking it.

Anyway, if you compare A(x-C)^2 + D to 1/(2[b-k])(x-a)^2 + .5(b+k) you should see A = 1/(2[b-k]), C = a and D = .5(b+k) so it is a much more complicated vertex form.  We can use this though.

Now, we know A = 1/(2[b-k]) and D = .5(b+k), so let's do a bit more work.

Starting with A = 1/(2[b-k]) you can use algebra to change it to 1/(2A) = b - k.  So you will be able to find b - k.  if we had one more equation with b and k we'd be able to get their exact values.  Well, we do.  D = .5(b + k) can eb written as 2D = b + k.  So now, if you have a parabla in vertex form you can take A and D and find b and k, where b is the y value of the focus and k is the directrix of the form y = k.  Of course the x value of the focus is just C.  Again, only upward and downward opening parabolas.

If you'd like me to run through an example let me know.

There is also a kind of shortcut method.  using a standard form you can rearrange your vertex form into (x-C)^2 = 4p(y-D) where 4p = 1/A or p = 1/[4A].  In this form you just need to memorize that the focus is (C, D + p) and the directrix is y = D - p.  if you prefer not to use the p then you have (C, D + 1/[4A]) and y = D - 1/[4A].  Once again, this is for vertical parabola and the horizontal ones are just a bit different.  You can us the same methods to find them though.

If there is anything you don't understand let me know and I can go more into it.  Or in general if you'd like more of an explanation on something.  This was kind of a complicated post.  I hope it helped.

EDIT

Thanks to muonsortsitout on reddit I've figured out how to explain that shhortcut method, with (x-C)^2 = 4p(y-D).

Starting back at 1/(2[b-k])(x-a)^2 + .5(b+k) you can rearrange things.

y = 1/(2[b-k])(x-a)^2 + .5(b+k)
(x - a)^2 = [2(y - (1/2)(b + k))](b - k)

If you set p = (b - k)/2 and q to (b + k)/2 you should see p is the distance in the y direction from the vertex to either the directrix or focus.  Or at least the value.  Specifically it is the distance to b.  -p is the distance to the directrix. q though is the average of the two values, which is the value right between them which is the y coordinate of the vertex.  Anyway this changes the equation.

(x - a)^2 = 4(y - q)p = 4p(y - q)  in what I wrote before q was D.  But yeah, that is how we this works.

the x coordinate of the  focus is of course a

the y coordinate is q + p since q is the y coordinate of the vertex and p is the distance between the vertex and the focus

then the directrix is y = q - p since you are going in the other direction from the vertex.

Again, let me know if something here didn't make sense.

it might help to rarrange (x-a)^2 = 4p(y-q) into y = 1/(4p) (x-a)^2 + q just to sww how things correspond.  We KNOW p is the distance from the vertex to either the

Question 9

How did the focus and directrix come to be?  It seems a bit too random and out of the blue.

Accepted Answer

The focus and directrix aren't just two random things. A cool property of this is that the distance from the focus to any point on the parabola will always be the same as the distance from that point on the parabola to the directrix.

Ever wonder how the headlights on your car work? It consists of a bulb at the focus of a parabolic mirror. When the light turns on, the light rays propagate spherically from the parabola's focus, when it comes into contact with mirror, the light will now be parallel to the ground. This is how your headlights concentrate the light to illuminate whats in front of your car. Hope this helps!

Algebra (all content)

Course: Algebra (all content) > Unit 17

Intro to focus & directrix

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Video transcript