Question 1

The situation where you are given, for example x=4 instead of y=4, was never covered in the videos.

Accepted Answer

Another consideration would be that when the directrix is a vertical line (x=k),  we are representing a parable that is faced to the right or to the left which is no longer a function unless its domain is limited to represent only one "arm" of the parable.

Question 2

couldn't you use the equation y= a(x-h)^2 +k and x=a(y-k)^2 +h, where a=1/4p?

Accepted Answer

In the equations  y=a(x-h)^2+k and x=a(y-k)^2+h where a=1/4p ,  (h,k ) represents vertex and p is the distance between focus and vertex .
But in the equations y=1/2(b-k) (x-a)^2+ (b+k)/2 and x=1/2(a-k ) (y-b)^2 +(a+k)/2 
 In the above  equations  (a,b) represents FOCUS not VERTEX . In the first one directrix is y=k and in second one directrix is x=k  . In the equations given by you in the question directrix in the first one is y=k-p and in second one it is x=h-p.
 Is it helpful ?

Question 3

what is the equation of a parabola having its focus at(3,4) and a directrix at X plus Y=1

Accepted Answer

The distance between (x,y) and (3,4) is √((x - 3)² + (y - 4)²). Similarly, the distance between (x,y) and the line x + y = 1 ⇔x + y - 1 = 0 is |x + y - 1| / √2.
√((x - 3)² + (y - 4)²) = |x + y - 1| / √2
(x - 3)² + (y - 4)² = (x + y - 1)² / 2
2x² - 12x + 18 + 2y² - 16y + 32 = x² + y² + 1 - 2x - 2y + 2xy
x² + y² - 10x - 14y - 2xy + 49 = 0

Question 4

In this page's exercise, the second problem says the parabola's directrix is at x=3, does this means this function is a horizontal one, like the inverse function of a traditional one? And if it is like that, should it have a domain so that there won't be the situation where one x will has two output?

Accepted Answer

I would be careful with the terminology. A parabola is only a function if it passes the Vertical Line Test, where you can test visually if an x input has more than 1 y input. In this case, it cannot be a function because each x has 2 y's (except the vertex). For this reason, they also cannot be true inverses of each other, because a function is only invertible if it is 1:1. A parabola is not 1:1, because two x inputs can yield the same output.
For example: y = x² , both -2 and 2 give y = 4. So if you were to invert this, the horizontal parabola cannot be a function; it wouldn't pass the VLT, because when x = 4, y = 2 _and_ -2. This is where, like you said, you would have to restrict the domain of the vertical parabola so that the inverse would exist.
A lengthy explanation, but I wanted things to make sense the best I could. Hope this helps!

Question 5

Why did you factor (y-5)^2 but not (x+2)^2 ?
in a problem in Khan Academy, I factored X but I got the wrong answer! the right answer was to factor the Y !!
I don't understand! 
here's the problem:
focus at (2,2), directrix x=8

Accepted Answer

If you need to find the X you multiply out the (x+2)^2.If you need to find the Y you factor out (y-5)^2.As is the example:

Write the equation for a parabola with a focus at(-2,5)and a directrix at 
x=3

x=? 
See,you need to find the X so you factor out the x.

Question 6

how do you solve equations that have a directrix that isn't vertical or horizontal?
for example: find the parabola with a focus (2,2) and a directrix y+x=-4

Accepted Answer

If you are familiar with matrix transformations there si a simplet method too. You still need to find the rotated angle and initial "unrotated" parabola of the form f(x) then treat it as a vector then multiply it by the matrix [, <-sin(t), cos(t)> Where t si the angle that was found the same way. If you multiply this out with matrix multiplication you get a new 2x1 matrix which describes the coordinates. This is actually a step in finding the formula from my first response, so if you want a formula without matrices it just leads to that.

Question 7

Find the Parabola with Focus (9,0) and Directrix y=-4

Accepted Answer

Any point (𝑥, 𝑦) on the parabola is equidistant to the focus and the directrix.

We can express these distances using the distance formula, and we get
√((𝑥 − 9)² + (𝑦 − 0)²) = √((𝑥 − 𝑥)² + (𝑦 − (−4))²)

Simplifying and squaring both sides gives us
(𝑥 − 9)² + 𝑦² = (𝑦 + 4)²

Expanding the squares and combining like terms we get
 𝑥² − 18𝑥 + 65 = 8𝑦

Then we divide both sides by 8 to get
𝑦 = (𝑥² − 18𝑥 + 65)∕8

Question 8

In the practice and this article many questions ask for x= but in the video _Sal Khan_ only went over how to find y=.

Accepted Answer

It's all pretty similar.  Follow the same steps from this video. https://www.khanacademy.org/math/geometry/xff63fac4:hs-geo-conic-sections/xff63fac4:hs-geo-parabola/v/equation-for-parabola-from-focus-and-directrix

Where the parabola is the line that is equidistant from a line x=h and the point (a,b) so the right side of the equation stays the same, but the left side has x-h.

Does that help?  Or I could show more explicitly what would happen.

Question 9

What would we do if y doesn't equal some simple number like -7 and equals something like x^2+7?

Accepted Answer

In that case, the curve would not be a parabola. The value of y(the equation of the directrix) must be dependent on x in a linear manner for the curve to be a parabola. y=-7, y=3.8, y=6x-8 and so on are permissible equations for the directrix, since they all are straight lines. However, you gave the example of a quadratic relation, which is not a straight line and hence cannot be the directrix of parabola.

High school geometry

Course: High school geometry > Unit 7

Parabola focus & directrix review

What are the focus and directrix of a parabola?

Parabola equation from focus and directrix

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