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cmy89 · Aug 11, 2008, 09:14 PM

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!

galactus · 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 $(y-k)^{2}=4p(x-h)$

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

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

$y^{2}-8y-7x+16=0$

galactus · Aug 14, 2008, 07:55 AM

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

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 · Jun 17, 2011, 06:30 AM

if your problem is x^2=8y what is your p vaule.

Unknown008 · Jun 17, 2011, 07:38 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 · Nov 28, 2011, 02:15 PM

equation of the parabola that satisfies the given conditions.Focus (6, 0), directrix x = 5

ebaines · Nov 28, 2011, 02:52 PM

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:
https://www.askmehelpdesk.com/mathem...ns-615388.html

maribel95 · May 20, 2012, 07:11 PM

what is the directrix and focus of y=x^2 and of x=(1/36) y^2?

ebaines · May 21, 2012, 10:14 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!

Teresaw3 · 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!