Lukecrayonz
  • Lukecrayonz
Hey @amistre64 :). Find the equation of the parabola with a vertex of (-1,2) and a focus of (-1,0)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
the distance from vertex to focus is imporant to know
Lukecrayonz
  • Lukecrayonz
That's all that is given..
Lukecrayonz
  • Lukecrayonz
Here's an example: http://screensnapr.com/v/LLWoVz.png

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More answers

amistre64
  • amistre64
|dw:1332548394602:dw|
amistre64
  • amistre64
you need to find the distance from the focus to the vertex with what is given
amistre64
  • amistre64
how would you find the distance between 2 given points?
Lukecrayonz
  • Lukecrayonz
d=xqrt(x2-x1)^2+(y2-y1)^2
Lukecrayonz
  • Lukecrayonz
sqrt* sorry.
Lukecrayonz
  • Lukecrayonz
So sqrt(-1+1)+(0-2), sqrt(-2), isqrt(2)
amistre64
  • amistre64
you need to practice your formula :) you should end up with a distance of 2
Lukecrayonz
  • Lukecrayonz
o.O Where did i go wrong?
Lukecrayonz
  • Lukecrayonz
oh. wow, i see
amistre64
  • amistre64
square your sums
Lukecrayonz
  • Lukecrayonz
i didnt square the answers haha
Lukecrayonz
  • Lukecrayonz
okay, so its -2^2, 4, sqrt(4)=2.
amistre64
  • amistre64
good now, the formula for a parabola in geometry is given as: 4ax = y^2 and there are 4 different versions of it depending on how the parabola works; but "a" is our distance from vertex to focus
amistre64
  • amistre64
so, we know this opens up towards the focus; which is situated below the vertex
Lukecrayonz
  • Lukecrayonz
I'm trying to do this based off the example, but its so strange..
Lukecrayonz
  • Lukecrayonz
http://screensnapr.com/v/ZX62B4.png
amistre64
  • amistre64
the example is using this format as well; but lets get the parabola oriented correctly, then we will move it to the correct vertex point
amistre64
  • amistre64
|dw:1332548943087:dw|
Lukecrayonz
  • Lukecrayonz
And the distance between them is two, since we solved for that, and p=2.
amistre64
  • amistre64
-x^2 = 4ay will set this up good; and a = 2
amistre64
  • amistre64
-x^2 = 8y yes
Lukecrayonz
  • Lukecrayonz
Oh, you do -x^2=4ay, my online book does p for a.
amistre64
  • amistre64
right
Lukecrayonz
  • Lukecrayonz
Okay, so our equation according to my book, is (x-h)^2=4p(y-k)
amistre64
  • amistre64
now that we have the parabola set up at the origin, we can move it to the given vertex
Lukecrayonz
  • Lukecrayonz
Since p is greater than zero, and it opens up.
Lukecrayonz
  • Lukecrayonz
Sorry, opens down haha, (x-h)^2=4p(y-k)
amistre64
  • amistre64
p measures distance at the moment, so it aint negative
amistre64
  • amistre64
the vector from v to f would give us distance and direction
amistre64
  • amistre64
so, if you want to keep in mind that f-v = distance and direction from v to f; then we can assign a sign to p
Lukecrayonz
  • Lukecrayonz
Didn't we solve for p though..?
amistre64
  • amistre64
we solved for the distance measured by p; but without specifying a direction in our workings, we cannot verify its direction (+ or -)
amistre64
  • amistre64
in other words, we have |p|
Lukecrayonz
  • Lukecrayonz
Makes sense, so what do we do from here?
amistre64
  • amistre64
a picture tells us the the parabola will open down; so if anything the distance and direction of p = -2, as long as we keep in mind that this is the vector from v to f, and not from v to d(the directix)
amistre64
  • amistre64
x^2 = 4(-2)y -> x^2=-8y now we "move" this parabola from the origin, to the given vertex point
amistre64
  • amistre64
(x-vx)^2 = -8(y-vy)
Lukecrayonz
  • Lukecrayonz
(-1,2)
Lukecrayonz
  • Lukecrayonz
y=-(1/8)(x+1)^2+2
amistre64
  • amistre64
yes, if we wnat this to look like a "usual" parabola equation
Lukecrayonz
  • Lukecrayonz
Just looked up the answer, wondering how they gt to there.
Lukecrayonz
  • Lukecrayonz
got*
Lukecrayonz
  • Lukecrayonz
"The distance between the vertex and the focus is p. p = sqrt [ ((-1+1)^2 +(2-0)^2] = sqrt(4) = 2 The focus lies below the vertex, so p=-2 and the parabola slopes downward. The vertex and focus lie on the line x=-1 which is vertical Therefore, the parabola is of the form (x-h)^2 = 4p(y-k) , (h,k) being the vertx (x+1)^2 = 4p(y-2) (x+1)^2=-8(y-2) y-2 = (-1/8) (x+1)^2 y = (-1/8)(x+1)^2 + 2"
amistre64
  • amistre64
that would be a good path to follow :)
Lukecrayonz
  • Lukecrayonz
Got it :D
Lukecrayonz
  • Lukecrayonz
Dont know why, this is one problem when you explained it, it actually got more complicating, haha. Usually I get it right away once you do it.
amistre64
  • amistre64
:) i blame the fall of rome
Lukecrayonz
  • Lukecrayonz
NOOOOOOOOOOOOOOOOOO, A WORD PROBLEM!
Lukecrayonz
  • Lukecrayonz
I BLAME THE CREATORS OF CALCULUS :'(
Lukecrayonz
  • Lukecrayonz
I'll make a new question..

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