# How do you find the vertex, focus, and directrix of parabolas?

I don't really understand this concept, so thorough examples would be much appreciated.

1. y^2=8x

2. x^2 + 8y + 16 = 0

... and
Find an equation of the parabola that satisfies the given conditions.
3. Focus (0,1), directrix y = 7

Thanks!

 galactus Posts: 2,272, Reputation: 1436 Ultra Member #2 Aug 12, 2008, 03:26 PM
Quote:
 Find an equation of the parabola that satisfies the given conditions. 3. Focus (0,1), directrix y = 7 Thanks!
The vertex of the parabola is halfway between the directrix and the focus.

So, the vertex is at (0,4)

The parabola opens down.

p is the distance (absolute value because distance is always positive) between the focus and the vertex and between the vertex and the directrix.

Therefore, p=3 because 7-4=3 and 4-1=3.

So, we have $(y-k)^{2}=4p(x-h)$

$(y-4)^{2}=4p(x-0)$

$(y-4)^{2}=7x$

$y^{2}-8y-7x+16=0$
 galactus Posts: 2,272, Reputation: 1436 Ultra Member #3 Aug 14, 2008, 08:55 AM
Quote:
 1. $y^{2}=8x$
|p| is the distance from the focus to the vertex and also the distance from the vertex to the

directrix. This one is of the form $x=a\cdot y^{2}$.

We can use

$y^{2}=4px$

or $x=\frac{1}{4p}y^{2}$

In this case, $x=\frac{1}{8}y^{2}$

$a=\frac{1}{8}$

Since $a=\frac{1}{4p}$, then p=2.

The vertex is at the origin and it opens to the right, so it has focus 2 units to the right of

the origin, F(2,0) and

directrix 2 units to the left of the origin at x=-2

Quote:
 2. $x^{2} + 8y + 16 = 0$
Can you do this one? Remember, you have the form: $(x-h)^{2}=4p(y-k)$

From your given equation, can you see what V(h,k) and p are?

I don't know if this matters or not. Have you even read, or even care about, anything I have posted to help you along? I posted the solutions to two of them so you would have an example to work from for other problems.
 joele12 Posts: 1, Reputation: 10 Junior Member #4 Jun 17, 2011, 07:30 AM
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #5 Jun 17, 2011, 08:38 AM

Put it in the form that galactus gave.

$x^2 = 8y$

$(x-h)^2 = 4p(y-k)$

So, you see clearly that h = 0. And if you expand;

$4p(y-k) = 4py - 4pk$

So, you see again that p = 2 and k = 0
 junemarie15 Posts: 2, Reputation: 10 Junior Member #6 Nov 28, 2011, 02:15 PM
equation of the parabola that satisfies the given conditions.Focus (6, 0), directrix x = 5
 ebaines Posts: 10,594, Reputation: 5809 Expert #7 Nov 28, 2011, 02:52 PM
Quote:
 Originally Posted by junemarie15 equation of the parabola that satisfies the given conditions.Focus (6, 0), directrix x = 5
Please do not double post questions. See:
 maribel95 Posts: 1, Reputation: 1 New Member #8 May 20, 2012, 08:11 PM
what is the directrix and focus of y=x^2 and of x=(1/36) y^2?
 ebaines Posts: 10,594, Reputation: 5809 Expert #9 May 21, 2012, 11:14 AM
For each of these first put the equation into standard form - either

$
y-y_1 = \frac 1 {4p} (x-x_1)^2
$

or

$
x-x_1 = \frac 1 {4p} (y-y_1)^2
$

Once you have this the vertex is at $(x_1, y_1)$, and the focus is a distance 'p' towards the "inside" of the parabola, and the directrix is a line that us distance 'p' from the cvertex towards the outside.

I'll do an example for you: consider

$
y = \frac 1 {12} (x-2)^2
$

The vertex is at (2,0), The value of 'p' is 3 (since 1/(4*3) = 1/12), and hence the focus is at (2,3). The directrix is a horizontal line a distance 3 below the vertex, or at y= -3.

Hope this helps!
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 Teresaw3 Posts: 1, Reputation: 1 New Member #10 Jun 19, 2012, 09:25 PM

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