Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, 9). I need help
@jhonyy9
y=a(x-h)^+k (h,k) is the vertex
since the vertex is at the origin, it is (0,0)
wait so what am I suppose to do?
@abb0t @jhonyy9 @koymoi
so the vertex is at the origin, and the focus is at (0, 9) what's the distance from the vertex to the focus?
9 right/
@jdoe0001
right so, notice for the focus to the origin, the "x" coordinate didn't change, the "y" did that means is a parabola moving upwards so the "focus form" of a parabola like so is \(\bf (x-h)^2=4p(y-k)\) (h, k) as the vertex p = distance from the vertex to the focus p > 0, opens upward p < 0, opens downward
ok that makes a little sense
so am I suppose to use that formula? how do I know what to plug in?
well, your vertex is at the origin, and your focal distance is 9
I'm not sure how doing that formula will give me one of my answer options, they seem like they don't match
origin (0,0)
0+p=9
now plug them in ;)
what is p suppose to be?
(h, k) as the vertex p = distance from the vertex to the focus p > 0, opens upward p < 0, opens downward
so p is 9
yes
so what's the next step after finding p?
plug the values in \(\bf (x-h)^2=4p(y-k)\)
actually if you know the vertex is at its origin, the formula could just be x^=4py or y^=4px
then like jdoe said you have to decide which one to use based on the value of p
I am seriously so confused
so x^=4py you know p=9
I am just suppose to have an answer in terms of y or y^2 apparently
plug it in you get x^=36y
after that what do I do? I mean my answer options are y = 1/36x2 y = 1/9x^2 y^2 = 9x y^2 = 36x
what did you get for the equation?
for the main one from earlier?
@jdoe0001
yes
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