[Jame145]

I need to find the focus , vertex, and the directrix of the parabola given by =(1/2)x^2+ 2x +4.
I don't understand how to find p.
I know the vertex is (-2,2).
h = -2. k=2.
I was told to use this (x-h)^2=4p(y-k)

[ebaines]

The objective is to convert with into the form . I'll show you an example, and then you can apply it to this specific problem.

Let's suppose that instead of the equation you were given you had . Steps are:

1. Get rid of the coefficient in front of the x^2 term, by dividing both sides by 3:



2. Now "complete the square" - this involves taking the coefficient in front of the 'x' term, halving it, then squaring that, and adding the result to both sides. For this example the coefficient of 'x' is 4/3; half of that is 2/3, and 2/3 squared is 4/9, so we add that to both sides:



Now notice that the part of the right hand side can be written as a square:



So the equation becomes:



3. Now we can rearrange to get the right hand side into the form (x-h)^2:



4. Finally rearrange the left hand side so that it's in the form 4p(y-k):



So the final equation is:




Now it's in the correct form of 4p(y-k) = (x-hk)^2, where
k = -1/6,
h = 2/3
p = 1/6

Hope this example helps. Now try these same steps with your problem, and let us know what you get for an answer.

[Jame145]

Yes thank you it was helpful. My answer was h=-2, y=2 , p=1/2

[ebaines]

Yes thank you it was helpful. My answer was h=-2, y=2 , p=1/2

Yes - if I understand what you mean the vertex is at (-2,2) and p = 1/2. So what do you get for the position of the focus, and the equation of the directriix?

[Jame145]

Yea. The focus was(-2, 2.5), and the directrix was(-2, 1.5)).

[ebaines]

yea. the focus was(-2, 2.5), and the directrix was(-2, 1.5)).

Correct!