| | | # How do you find the vertex, focus, and directrix of parabolas?
Asked Aug 11, 2008, 09:14 PM
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** 9 Answers** | I don't really understand this concept, so thorough examples would be much appreciated. :D
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! |
9 Answers
| Ultra Member | |
Aug 12, 2008, 02:26 PM
| |
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 | | | Ultra Member | |
Aug 14, 2008, 07:55 AM
| |
1. |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 .
We can use
or
In this case,
Since , 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**
2. Can you do this one? Remember, you have the form:
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. | | | New Member | |
Jun 17, 2011, 06:30 AM
| | if your problem is x^2=8y what is your p vaule. | | | Uber Member | |
Jun 17, 2011, 07:38 AM
| |
Put it in the form that galactus gave.
So, you see clearly that h = 0. And if you expand;
So, you see again that p = 2 and k = 0 | | | New Member | |
Nov 28, 2011, 01:15 PM
| | equation of the parabola that satisfies the given conditions.Focus (6, 0), directrix x = 5 | | | New Member | |
May 20, 2012, 07:11 PM
| | what is the directrix and focus of y=x^2 and of x=(1/36) y^2? | | | Expert | |
May 21, 2012, 10:14 AM
| | For each of these first put the equation into standard form - either
or
Once you have this the vertex is at , 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
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! | | | New Member | |
Jun 19, 2012, 08:25 PM
| | What is the directrix of (y-(-2))^2=3(x-(-1))? I do not understand how to find the answer for this please help! | | Question Tools | Search this Question | | | ### Add your answer here. ##
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