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.
the distance from vertex to focus is imporant to know
That's all that is given..
Here's an example: http://screensnapr.com/v/LLWoVz.png
you need to find the distance from the focus to the vertex with what is given
how would you find the distance between 2 given points?
So sqrt(-1+1)+(0-2), sqrt(-2), isqrt(2)
you need to practice your formula :) you should end up with a distance of 2
o.O Where did i go wrong?
oh. wow, i see
square your sums
i didnt square the answers haha
okay, so its -2^2, 4, sqrt(4)=2.
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
so, we know this opens up towards the focus; which is situated below the vertex
I'm trying to do this based off the example, but its so strange..
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
And the distance between them is two, since we solved for that, and p=2.
-x^2 = 4ay will set this up good; and a = 2
-x^2 = 8y yes
Oh, you do -x^2=4ay, my online book does p for a.
Okay, so our equation according to my book, is (x-h)^2=4p(y-k)
now that we have the parabola set up at the origin, we can move it to the given vertex
Since p is greater than zero, and it opens up.
Sorry, opens down haha, (x-h)^2=4p(y-k)
p measures distance at the moment, so it aint negative
the vector from v to f would give us distance and direction
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
Didn't we solve for p though..?
we solved for the distance measured by p; but without specifying a direction in our workings, we cannot verify its direction (+ or -)
in other words, we have |p|
Makes sense, so what do we do from here?
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)
x^2 = 4(-2)y -> x^2=-8y now we "move" this parabola from the origin, to the given vertex point
(x-vx)^2 = -8(y-vy)
yes, if we wnat this to look like a "usual" parabola equation
Just looked up the answer, wondering how they gt to there.
"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"
that would be a good path to follow :)
Got it :D
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.
:) i blame the fall of rome
NOOOOOOOOOOOOOOOOOO, A WORD PROBLEM!
I BLAME THE CREATORS OF CALCULUS :'(
I'll make a new question..