anonymous
  • anonymous
Find the standard form of the equation of the parabola with a focus at (-4, 0) and a directrix at x = 4. Could someone please help?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
wintersuntime
  • wintersuntime
Write an equation for the parabola with focus (4,0) and directrix y=2. Thanks! .. Standard form of parabola: (x-h)^2=4p(y-k), with (h,k) being the (x,y) coordinates of the vertex Given parabola opens downward with axis of symmetry at x=4. Vertex at (4,1). p=1 Equation of parabola: (x-4)^2=-4(y-1) (ans) see the graph below as a visual check on the answer .. y=1-(x-4)^2/4
anonymous
  • anonymous
I'm so confused I'm sorry @wintersuntime
wintersuntime
  • wintersuntime
Are you trying to graph it ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

wintersuntime
  • wintersuntime
And it's okay I love helping others
anonymous
  • anonymous
No. These are the answer I am given: A. y^2 = -8x B. 16y = x^2 C. y = -1/16x^2 D. x = -1/16y^2
wintersuntime
  • wintersuntime
What do you think the answer will be ?
anonymous
  • anonymous
I am thinking C. @wintersuntime
wintersuntime
  • wintersuntime
why ?
anonymous
  • anonymous
Honestly, I'm not sure. I just thought that standard form looked most similar to that choice. @wintersuntime
anonymous
  • anonymous
its not C
anonymous
  • anonymous
Could you explain how to find the correct answer then? I am so lost. @saseal
anonymous
  • anonymous
|dw:1438996328137:dw|
anonymous
  • anonymous
the parabola is opening down because the focus is below the directrix
anonymous
  • anonymous
But the parabola's focus is at -4, 0. So would that mean the parabola's focus would be shifted to the left 4? Not down 4?
anonymous
  • anonymous
no, focus has nothing to do with shifting
anonymous
  • anonymous
Yes, but you put the point at (0, -4) it looks like.
anonymous
  • anonymous
lel i made a mistake at the drawing suppose to be (0,p)
anonymous
  • anonymous
Ok, but I'm still confused as to what the standard form would look like?
anonymous
  • anonymous
the standard form flavor looks like this \[x^2=4py\] or \[y^2=4px\]
anonymous
  • anonymous
depends on where your parabola opens out
anonymous
  • anonymous
Ok so since the parabola opens down, would it be y^2 = 4px?
anonymous
  • anonymous
|dw:1438996750393:dw|
anonymous
  • anonymous
|dw:1438996883193:dw|
anonymous
  • anonymous
so its not y^2
anonymous
  • anonymous
Oh ok, so it would be in the form of x^2 = 4py?
anonymous
  • anonymous
yes
anonymous
  • anonymous
So the answer then would be B? @saseal
Loser66
  • Loser66
Tip: no need to consider whether p >0 or p <0, if you see the focus, just plug that p in For example, on your problem, focus is (0,-4) , hence plug -4 in the form x^2 = 4 p y x^2 = 4*(-4) y.
Loser66
  • Loser66
then isolate y to get the form.
anonymous
  • anonymous
should be x^2 = -16y
anonymous
  • anonymous
That's not the answer. I just checked.
anonymous
  • anonymous
I think it's y = -1/16x^2
Loser66
  • Loser66
surely not \(x^2 = -16y\) divided both sides by -16, you get y = -(1/16)x^2
anonymous
  • anonymous
Ok I was correct. Thank you!
anonymous
  • anonymous
I wish I could give you all medals.
Loser66
  • Loser66
give it to saseal, I have more than 3 thousands of it. hehehe.
anonymous
  • anonymous
lol
anonymous
  • anonymous
Could you guys just check a couple for me? I'm really confused on these, and have them done, just want to make sure they're correct.
anonymous
  • anonymous
alright
anonymous
  • anonymous
Find the vertex, focus, directrix, and focal width of the parabola. x = 3^y2 Would the answer be: Vertex: (0,0) Focus: (1/12,0) Focus Width: .333 or 1/12 Directrix: -1/12
anonymous
  • anonymous
\[x=3y^2\]\[y^2=\frac{ 1 }{ 3 }x\]\[y^2 = 4(\frac{ 1 }{ 12 }) x\]
anonymous
  • anonymous
Ok...? So I'm incorrect?
anonymous
  • anonymous
focus width is wrong and directrix usually they write it as directrix = x = -1/12 so people know which axis it is on
anonymous
  • anonymous
Oh ok sorry. But my choice for the vertex, focus, and directrix to all remain the same, the focal width has to be .33
anonymous
  • anonymous
yes
anonymous
  • anonymous
focal width aka latus rectum = 4p
anonymous
  • anonymous
That's not one of my choices. I think they wanted you to solve for the width.
anonymous
  • anonymous
|dw:1438997923182:dw|
anonymous
  • anonymous
Ok thanks!
anonymous
  • anonymous
For: A building has an entry the shape of a parabolic arch 74 ft high and 28 ft wide at the base as shown below. Would my answer be: y = −37/98x^2
anonymous
  • anonymous
|dw:1438998052704:dw|
anonymous
  • anonymous
Ok
anonymous
  • anonymous
so its definitely p<0
anonymous
  • anonymous
So what would that mean for the ending answer?
Loser66
  • Loser66
To this kind of problem, you need derive from the standard form Vertex (0,74), Two points are (-14,0) and (14,0) ok?
anonymous
  • anonymous
Ok
Loser66
  • Loser66
Vertex form is \(y = a(x-h)^2 +k\) \(y = ax^2 + 74\)
anonymous
  • anonymous
|dw:1438998433802:dw|
anonymous
  • anonymous
always better t o draw it out haha
Loser66
  • Loser66
Now, pick one of the point, I pick (14,0) , that is x =14 , y =0 and plug them in \(0= a (14)^2 +74\) \(a(14)^2 = -74\) \(a = \dfrac{-74}{196}\)
Loser66
  • Loser66
Now, replace back \(y = \dfrac{-74}{196}x^2\) @saseal they didn't say the base of the parabola is the line passed through focus. You cannot apply the form of focus, directrix here.
Loser66
  • Loser66
@jdoherty Got what I meant?
anonymous
  • anonymous
Ok, so my answer would be correct then. I do understand. (:
anonymous
  • anonymous
yea ikr thats why i didnt use anything to do with focus
Loser66
  • Loser66
your answer is (-37/98)x^2, that is not correct.
anonymous
  • anonymous
But I just simplified the fraction, dividing both the top and bottom by 2... And I just have one more question. Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7). Those still are the most confusing to me.
anonymous
  • anonymous
x^2 = 4 (-7)y
Loser66
  • Loser66
When they give you the focus or directrix, they want you to define the form of the parabola. You know that the focus is ALWAYS inside of the parabola, right? hence ,your parabola is up or down?
anonymous
  • anonymous
So it would be: y = -1/28 x^2?
anonymous
  • anonymous
@jdoherty refer to the drawings above to see where the parabola opens up
anonymous
  • anonymous
I did refer to that, but based on the equation, I got my answer of y = -1/28x^2
anonymous
  • anonymous
@saseal
anonymous
  • anonymous
yea?
anonymous
  • anonymous
so thats correct
anonymous
  • anonymous
sorry i fell asleep

Looking for something else?

Not the answer you are looking for? Search for more explanations.