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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

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).
A parabola that opens up. Above the vertex of the parabola is a point labeled focus. Below the parabola is a horizontal line labeled directirix. On the parabola, there are three points at random locations. Each point has a line segment that attaches to the focus and a line segment that attaches to the directrix. The pairs of segments that leave a point on the parabola have equal distances.

## 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:
\begin{aligned} \sqrt{(y-3)^2} &= \sqrt{(x+2)^2+(y-5)^2} \\\\ (y-3)^2 &= (x+2)^2+(y-5)^2 \\\\ \blueD{y^2}-6y\goldD{+9} &= (x+2)^2\blueD{+y^2}\maroonD{-10y}+25 \\\\ -6y\maroonC{+10y}&=(x+2)^2+25\goldD{-9} \\\\ 4y&=(x+2)^2+16 \\\\ y&=\dfrac{(x+2)^2}{4}+4\end{aligned}

Problem 1
• Current
Write the equation for a parabola with a focus at left parenthesis, 6, comma, minus, 4, right parenthesis and a directrix at y, equals, minus, 7.
y, equals

Want to try more problems like this? Check out this exercise.

## 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? •   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.
• The situation where you are given, for example x=4 instead of y=4, was never covered in the videos. • 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 • 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.
• what is the equation of a parabola having its focus at(3,4) and a directrix at X plus Y=1 • 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
• 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? • 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!
• In the practice and this article many questions ask for x= but in the video Sal Khan only went over how to find y=. • I still do not know what a focus and a directrix is? Can someone explain to me? • 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.
• Find the Parabola with Focus (9,0) and Directrix y=-4 • 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  