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

Lesson 7: Focus and directrix of a parabola

# Intro to focus & directrix

A parabola is the set of all points equidistant from a point (called the "focus") and a line (called the "directrix"). See this video to learn more about this.

## Want to join the conversation?

• How to prove that every point in the parabolic curve will have the same distance to the focus and directrix?
• It's defined that way. Definitions don't require proof.
• What is the importance of having a directrix? Are there any applications of it in real life?
• 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.
• 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`?
• 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.
• 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
• 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.
• 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?
• 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.
• What if the directrix of a parabola is vertical? Would that make a horizontal parabola?
• Yes, we would get a horizontal parabola in that case.
• Could a parabolic equation be written for a slanted directrix?
• 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.
• how do you find focus and directrix of a parabola
• 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
• Is the vertex of a parabola always halfway between the focus and the directrix, or was that just an example Sal made?
• 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
• How can we prove that the equation ax^2 + bx + c creates a parabola? How can we prove that slicing a cone also creates a parabola?
• Slicing a cone is really just visual. If you look at all conic sections they are literally giving every possible way to slide a cone, or more specifically two cones positioned tip to tip, and describing what happens with each method.

As to prove that the standard form works the solution is to start with the most basic form of a parabola and work from there.

x^2

Now, before anything else you should eb familiar with graph transformations. Moving it left and right, or up and down, or manipulating how wide/ skinny it is. If you are familiar with graph transformations this should look familiar. A*f(B(x - C)) + D Where A is the vertical stretch, that is you multiply the y value of all coordinates by A except for where x=C. The point (C,D) is the only point that will not experience any stretches or shrinks. B is the horizontal shrink, so you divide all x values by B. c is the horizontal shift, so ALL x vales move C to the right, or if C is negative (so it would look like (x - -C) or (x + C) )you move to the left. D is the vertical shift so you move up D unless D is negative in which case you move down.

To write these transformations for x^2 s would look like A(B(x - C))^2 + D Although you could factor out the B to make it AB^2 * (x - C)^2 + D So AB^2 will give the effect of the horizontal and vertical stretch and shrink. This works because 8x^2, where x squared is getting stretched vertically by 8 is the same as 2(2x)^2 where x^2 is getting a vertical stretch by 2 and a horizontal shrink by 2. This may not be obvious, but if you use a point x = a you get a coordinate pair of (a, 8a^2) meanwhile if you do the same for 8x^2 (a, 8a^2). or if you'd rather use real numbers you could solve for 3. in 2(2x)^2 for x=3 you get (3, 72) meanwhile 8x62 gets you (3, 72). So hopefully you can understand 2(2x)^2 is the same as 8x^2.

NOW! I am going to say AB^2 is just A for ease. This leaves us with A(x - C)^2 + D. If you expand it you get Ax^2 - 2ACx + AC^2 + D. This leaves us witht he standard form where A = a, -2AC = b and AC^2 + D = c. So ax^2 + bx + c is just a rewritten version of x^2 with its transformations.

Does this make sense? If not you can agan think of it with real numbers. If you put actual numbers in for A, C and D then expand it you will get something in the form of ax^2 + bx + c where a, b and c are actual numbers.

If you are unaware there is also a way to change ax^2 + bx + c into A(x - C) + D, called changing from standard form to vertex form.

Let me know if you do need more explanation.