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## High school geometry

### Course: High school geometry > Unit 7

Lesson 4: Focus and directrix of a parabola# Parabola focus & directrix review

CCSS.Math:

Review your knowledge of the focus and directrix of parabolas.

## What are the focus and directrix of a parabola?

Parabolas are commonly known as the graphs of quadratic functions. They can also be viewed as the set of all points whose distance from a certain point (the

**focus**) is equal to their distance from a certain line (the**directrix**).*Want to learn more about focus and directrix of a parabola? Check out this video.*

## Parabola equation from focus and directrix

Given the focus and the directrix of a parabola, we can find the parabola's equation. Consider, for example, the parabola whose focus is at left parenthesis, minus, 2, comma, 5, right parenthesis and directrix is y, equals, 3. We start by assuming a general point on the parabola left parenthesis, x, comma, y, right parenthesis.

Using the distance formula, we find that the distance between left parenthesis, x, comma, y, right parenthesis and the focus left parenthesis, minus, 2, comma, 5, right parenthesis is square root of, left parenthesis, x, plus, 2, right parenthesis, squared, plus, left parenthesis, y, minus, 5, right parenthesis, squared, end square root, and the distance between left parenthesis, x, comma, y, right parenthesis and the directrix y, equals, 3 is square root of, left parenthesis, y, minus, 3, right parenthesis, squared, end square root. On the parabola, these distances are equal:

*Want to learn more about finding parabola equation from focus and directrix? Check out this video.*

## Want to join the conversation?

- couldn't you use the equation y= a(x-h)^2 +k and x=a(y-k)^2 +h, where a=1/4p?(21 votes)
- 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 ?(58 votes)

- The situation where you are given, for example x=4 instead of y=4, was never covered in the videos.(42 votes)
- 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.(7 votes)

- 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(4 votes)- 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.(12 votes)

- what is the equation of a parabola having its focus at(3,4) and a directrix at X plus Y=1(6 votes)
- 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(6 votes)

- 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?(5 votes)
- 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!(5 votes)

- In the practice and this article many questions ask for x= but in the video
*Sal Khan*only went over how to find y=.(3 votes)- 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.(5 votes)

- I still do not know what a focus and a directrix is? Can someone explain to me?(2 votes)
- A focus and directrix are just a point and a line, respectively. We can define a parabola as the set of all points that are equidistant from the focus and the directrix.

Given any line and a point not on it, we can find one, unique parabola that is always equidistant from them, i.e. one parabola that has the given point and line as its focus and directrix.

Conversely, given any parabola, we can find its focus and directrix. It can be helpful to think of the focus and directrix as a 'part' of the parabola, in the same way that the center and radius of a circle are 'part' of the circle; they aren't actually on the curve, but they carry all of the information we need to construct it.(5 votes)

- Find the Parabola with Focus (9,0) and Directrix y=-4(2 votes)
- 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(4 votes)

- 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(3 votes)- 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 <x, f(x)> then multiply it by the matrix [<cos(t), sin(t)>, <-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 <x*cos(t) - f(x)*sin(t), x*sin(t) + f(x)*cos(t)> 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.(2 votes)

- Is there an easier way to remember the distance formula?(3 votes)
- If you relate the distance formula to the slope formula m=(y2-y1)/(x2-x1) and Pythagorean theorem (c= sqrt(a^2+b^2)) with distance formula being sqrt((x2-x1)^2+(y2-y1)^2) maybe that would help.

If you have two points or a table, draw line between 2 y values and subtract on top and draw line between 2 x values on bottom and subtract. If linear, slope = top/bottom and distance = sqrt(top^2+bottom^2)(3 votes)